calculus i - lecture 23 fundamental theorem of calculusgerald/math220d/lec23.pdfsection 5.3 -...
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![Page 1: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/1.jpg)
Calculus I - Lecture 23Fundamental Theorem of Calculus
Lecture Notes:http://www.math.ksu.edu/˜gerald/math220d/
Course Syllabus:http://www.math.ksu.edu/math220/spring-2014/indexs14.html
Gerald Hoehn (based on notes by T. Cochran)
April 16, 2014
![Page 2: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/2.jpg)
Section 5.3 - Fundamental Theorem of Calculus I
We have seen two types of integrals:
1. Indefinite:
∫f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x).
2. Definite:
∫ b
af (x) dx = signed area bounded by f (x) over [a, b].
Theorem (Fundamental Theorem of Calculus I)
Let f (x) be a continuous function on [a, b]. Then∫ b
af (x) dx = F (x)
∣∣∣∣ba
:= F (b)− F (a)
where F (x) is an antiderivative of f (x).
Note: The result is independent of the chosen antiderivative F (x).
![Page 3: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/3.jpg)
Section 5.3 - Fundamental Theorem of Calculus I
We have seen two types of integrals:
1. Indefinite:
∫f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x).
2. Definite:
∫ b
af (x) dx = signed area bounded by f (x) over [a, b].
Theorem (Fundamental Theorem of Calculus I)
Let f (x) be a continuous function on [a, b]. Then∫ b
af (x) dx = F (x)
∣∣∣∣ba
:= F (b)− F (a)
where F (x) is an antiderivative of f (x).
Note: The result is independent of the chosen antiderivative F (x).
![Page 4: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/4.jpg)
Section 5.3 - Fundamental Theorem of Calculus I
We have seen two types of integrals:
1. Indefinite:
∫f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x).
2. Definite:
∫ b
af (x) dx = signed area bounded by f (x) over [a, b].
Theorem (Fundamental Theorem of Calculus I)
Let f (x) be a continuous function on [a, b]. Then∫ b
af (x) dx = F (x)
∣∣∣∣ba
:= F (b)− F (a)
where F (x) is an antiderivative of f (x).
Note: The result is independent of the chosen antiderivative F (x).
![Page 5: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/5.jpg)
Section 5.3 - Fundamental Theorem of Calculus I
We have seen two types of integrals:
1. Indefinite:
∫f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x).
2. Definite:
∫ b
af (x) dx = signed area bounded by f (x) over [a, b].
Theorem (Fundamental Theorem of Calculus I)
Let f (x) be a continuous function on [a, b]. Then∫ b
af (x) dx = F (x)
∣∣∣∣ba
:= F (b)− F (a)
where F (x) is an antiderivative of f (x).
Note: The result is independent of the chosen antiderivative F (x).
![Page 6: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/6.jpg)
Section 5.3 - Fundamental Theorem of Calculus I
We have seen two types of integrals:
1. Indefinite:
∫f (x) dx = F (x) + C
where F (x) is an antiderivative of f (x).
2. Definite:
∫ b
af (x) dx = signed area bounded by f (x) over [a, b].
Theorem (Fundamental Theorem of Calculus I)
Let f (x) be a continuous function on [a, b]. Then∫ b
af (x) dx = F (x)
∣∣∣∣ba
:= F (b)− F (a)
where F (x) is an antiderivative of f (x).
Note: The result is independent of the chosen antiderivative F (x).
![Page 7: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/7.jpg)
We have three ways of evaluating definite integrals:
1. Use of area formulas if they are available.
(This is what we did last lecture.)
2. Use of the Fundamental Theorem of Calculus (F.T.C.)
3. Use of the Riemann sum limn→∞
∑ni=1 f (xi )∆x
(This we will not do in this course.)
![Page 8: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/8.jpg)
We have three ways of evaluating definite integrals:
1. Use of area formulas if they are available.
(This is what we did last lecture.)
2. Use of the Fundamental Theorem of Calculus (F.T.C.)
3. Use of the Riemann sum limn→∞
∑ni=1 f (xi )∆x
(This we will not do in this course.)
![Page 9: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/9.jpg)
We have three ways of evaluating definite integrals:
1. Use of area formulas if they are available.
(This is what we did last lecture.)
2. Use of the Fundamental Theorem of Calculus (F.T.C.)
3. Use of the Riemann sum limn→∞
∑ni=1 f (xi )∆x
(This we will not do in this course.)
![Page 10: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/10.jpg)
We have three ways of evaluating definite integrals:
1. Use of area formulas if they are available.
(This is what we did last lecture.)
2. Use of the Fundamental Theorem of Calculus (F.T.C.)
3. Use of the Riemann sum limn→∞
∑ni=1 f (xi )∆x
(This we will not do in this course.)
![Page 11: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/11.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 12: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/12.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:
∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 13: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/13.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx
= A− B =1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 14: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/14.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B
=1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 15: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/15.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1
= 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 16: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/16.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2
=3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 17: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/17.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)
∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 18: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/18.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 19: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/19.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 20: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/20.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 21: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/21.jpg)
Example: Evaluate
∫ 3
0(2− x) dx using the first two methods.
Solution:
1) Areas:∫ 3
0(2− x) dx = A− B =
1
2· 2 · 2− 1
2· 1 · 1 = 2− 1
2=
3
2
2) F.T.C. (No graph required)∫ 3
0(2− x) dx = 2x − x2
2︸ ︷︷ ︸antideriv.
∣∣∣∣30
= (2 · 3− 32
2)− (2 · 0− 02
2)
= 6− 9
2=
3
2
![Page 22: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/22.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)=
3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 23: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/23.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)=
3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 24: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/24.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)=
3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 25: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/25.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)
=3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 26: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/26.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)=
3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 27: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/27.jpg)
Example: Evaluate using the F.T.C.∫ 8
1
(13√
x− 5
x
)dx .
Solution:∫ 8
1
(13√
x− 5
x
)dx =
∫ 8
1
(x−1/3 − 5x−1
)dx
=
(x2/3
2/3− 5 ln |x |
)∣∣∣∣∣8
1
=
(3
2· 82/3 − 5 ln 8
)−(
3
2− 5 ln 1
)=
3
2· 4− 5 ln 8− 3
2
=9
2− 5 ln 8
![Page 28: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/28.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)= 0 +
1
3=
1
3
![Page 29: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/29.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)= 0 +
1
3=
1
3
![Page 30: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/30.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)
= 0 +1
3=
1
3
![Page 31: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/31.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)= 0 +
1
3
=1
3
![Page 32: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/32.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)= 0 +
1
3=
1
3
![Page 33: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/33.jpg)
Example: Evaluate using the F.T.C.∫ π/2
0sin(3x) dx .
Solution:∫ π/2
0sin(3x) dx = − cos(3x) · 1
3
∣∣∣∣π/20
= −1
3cos(3π
2
)−(− 1
3cos(0)
)= 0 +
1
3=
1
3
![Page 34: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/34.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8 =
16
3
![Page 35: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/35.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8 =
16
3
![Page 36: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/36.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8 =
16
3
![Page 37: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/37.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8 =
16
3
![Page 38: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/38.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8
=16
3
![Page 39: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/39.jpg)
Example: Find the area of the region below.
Solution:
A =
∫ 4
0
√x dx =
∫ 4
0x1/2 dx
=x3/2
3/2
∣∣∣∣40
=2
3· 43/2 − 0
=2
3· 8 =
16
3
![Page 40: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/40.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 41: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/41.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 42: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/42.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 43: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/43.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 44: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/44.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 45: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/45.jpg)
Section 5.4 - Fundamental Theorem of Calculus IILet f (t) be a continuous function on [a, b].
Let x be a point with a < x < b.
Let A(x) =
∫ x
af (t) dt = Signed Area bounded by f (t) over [a, x ].
Goal: Find the rate that the area A(x) increases or decreases, thatis, find dA
dx .
Let dx = infinitesimal change in x .
dA = resulting change in the area
dA = height × base = f (x) · dx .
Thus:dA
dx=
d
dx
(∫ x
af (t) dt
)= f (x).
![Page 46: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/46.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 47: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/47.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 48: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/48.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 49: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/49.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 50: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/50.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 51: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/51.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt
=1
u2 + 3, by F.T.C. II.
![Page 52: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/52.jpg)
Theorem (Fundamental Theorem of Calculus II)
Let f (x) be a continuous function on [a, b]. Then for any x in(a, b) we have
d
dx
∫ x
af (t) dt = f (x)
where f (x) is the evaluation of f (t) at x.
“The derivative of the integral of a function is the function.”
Example: Findd
dx
∫ x
2et · cos(5t) dt
Solution:d
dx
∫ x
2et · cos(5t) dt = ex cos(5x), by F.T.C. II.
Example: Findd
du
∫ u
−3
1
t2 + 1dt
Solution:d
du
∫ u
−3
1
t2 + 1dt =
1
u2 + 3, by F.T.C. II.
![Page 53: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/53.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 54: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/54.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 55: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/55.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 56: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/56.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 57: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/57.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 58: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/58.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)
= − sin x2
![Page 59: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/59.jpg)
Flipping the limits of integration
DefinitionLet f (x) be a continuous function on [a, b]. Then∫ a
bf (x) dx := −
∫ b
af (x) dx.
It follows thatd
dx
∫ a
xf (t) dt = −f (x).
When variable is lower limit insert (-) sign.
Example: Findd
dx
∫ 5
xsin t2 dt
Solution:d
dx
∫ 5
xsin t2 dt
=d
dx
(−∫ x
5sin t2 dt
)= − sin x2
![Page 60: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/60.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 61: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/61.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 62: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/62.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 63: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/63.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 64: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/64.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 65: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/65.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx
= f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 66: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/66.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 67: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/67.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt
= f (x).
![Page 68: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/68.jpg)
Concept of the “dummy” variable
a) Let F (x) =
∫ x
at2 dt =
t3
3
∣∣∣∣xa
=x3
3− a3
3
t is a dummy variable
b) Let F (x) =
∫ x
au2 du =
u3
3
∣∣∣∣xa
=x3
3− a3
3
u is a dummy variable.
We see that F (x) = G (x). The name of the dummy variable playsno role for the value of the integral.
Example: Find:
a)d
dt
∫ t
3f (x) dx = f (t).
b)d
dx
∫ x
3f (t) dt = f (x).
![Page 69: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/69.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 70: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/70.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 71: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/71.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 72: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/72.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 73: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/73.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 74: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/74.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 75: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/75.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2
=2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 76: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/76.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 77: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/77.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 78: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/78.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt
=d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 79: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/79.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 80: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/80.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3
= 3x2 cos x6 − cos x2
![Page 81: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/81.jpg)
An extension of the F.T.C. II
d
dx
∫ g(x)
af (t) dt = f (g(x)) · g ′(x)
Indeed, let F (x) =∫ xa f (t) dt. Then the chain rule gives:
d
dxF (g(x)) = F ′(g(x)) · g ′(x) = f (g(x)) · g ′(x).
Example: Findd
dx
∫ x2
2
dt√t + 1
.
Solution:d
dx
∫ x2
2
dt√t + 1
=1√
x2 + 1· d
dxx2 =
2x
x + 1
Example: Findd
dx
∫ x3
xcos t2 dt.
Solution:d
dx
∫ x3
xcos t2 dt =
d
dx
∫ a
xcos t2 dt +
d
dx
∫ x3
acos t2 dt
= − cos x2 + cos((x3)2
)· d
dxx3 = 3x2 cos x6 − cos x2
![Page 82: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/82.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 83: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/83.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 84: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/84.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 85: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/85.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 86: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/86.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 87: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/87.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 88: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/88.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 89: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/89.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I
![Page 90: Calculus I - Lecture 23 Fundamental Theorem of Calculusgerald/math220d/lec23.pdfSection 5.3 - Fundamental Theorem of Calculus I We have seen two types of integrals: 1. Inde nite: Z](https://reader034.vdocuments.net/reader034/viewer/2022051814/6036f0216eee6202770604bb/html5/thumbnails/90.jpg)
We have seen two ways to find an antiderivative of f (x):
1. Use our known formulas for derivatives and work backwards:
Let F (x) be such that F ′(x) = f (x).
2. Use a definite integral: Let
A(x) =
∫ x
af (t) dt.
Then A′(x) = f (x) and A(a) = 0.
We have also seen that any two antiderivatives must differ by aconstant. Thus:
A(x) = F (x) + C .
Let us find C :
A(a) = 0 = F (a) + C ⇒ C = −F (a).
Thus: A(x) = F (x)− F (a).
Therefore:
∫ b
af (t) dt = A(b) = F (b)− F (a).
This is the F.T.C. I