lesson 25: evaluating definite integrals (section 041 slides)
DESCRIPTION
A remarkable theorem about definite integrals is that they can be calculated with antiderivatives.TRANSCRIPT
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Section 5.3Evaluating Definite Integrals
V63.0121.041, Calculus I
New York University
December 6, 2010
Announcements
I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13: Section 5.5I ”Monday,” December 15: Review and Movie Day!I Monday, December 20, 12:00–1:50pm: Final Exam
. . . . . .
. . . . . .
Announcements
I Today: Section 5.3I Wednesday: Section 5.4I Monday, December 13:Section 5.5
I ”Monday,” December 15:Review and Movie Day!
I Monday, December 20,12:00–1:50pm: Final Exam
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 2 / 41
. . . . . .
Objectives
I Use the EvaluationTheorem to evaluatedefinite integrals.
I Write antiderivatives asindefinite integrals.
I Interpret definite integralsas “net change” of afunction over an interval.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 3 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 4 / 41
. . . . . .
The definite integral as a limit
DefinitionIf f is a function defined on [a,b], the definite integral of f from a to bis the number ∫ b
af(x)dx = lim
n→∞
n∑i=1
f(ci)∆x
where ∆x =b− an
, and for each i, xi = a+ i∆x, and ci is a point in[xi−1, xi].
TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b
af(x) dx exists and is the same for any choice of ci.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 5 / 41
. . . . . .
Notation/Terminology
∫ b
af(x)dx
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 6 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=
14
(4
65/64+
473/64
+4
89/64+
4113/64
)
=150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=
14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 7 / 41
. . . . . .
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a,b] and c a constant. Then
1.∫ b
ac dx = c(b− a)
2.∫ b
a[f(x) + g(x)] dx =
∫ b
af(x)dx+
∫ b
ag(x)dx.
3.∫ b
acf(x)dx = c
∫ b
af(x)dx.
4.∫ b
a[f(x)− g(x)] dx =
∫ b
af(x)dx−
∫ b
ag(x)dx.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 8 / 41
. . . . . .
More Properties of the Integral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx∫ a
af(x)dx = 0
This allows us to have
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 9 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
.
∫ c
bf(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x)dx
.
∫ c
bf(x)dx
.
∫ c
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ b
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ c
bf(x)dx =
−∫ b
cf(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Illustrating Property 5
Theorem
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ c
bf(x)dx =
−∫ b
cf(x)dx
.
∫ c
af(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 10 / 41
. . . . . .
Definite Integrals We Know So Far
I If the integral computes anarea and we know thearea, we can use that. Forinstance,∫ 1
0
√1− x2 dx =
π
4
I By brute force wecomputed∫ 1
0x2 dx =
13
∫ 1
0x3 dx =
14
..x
.
y
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 11 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].
6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].
6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].
6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 12 / 41
. . . . . .
Estimating an integral with inequalities
Example
Estimate∫ 2
1
1xdx using Property 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1we have
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
. . . . . .
Estimating an integral with inequalities
Example
Estimate∫ 2
1
1xdx using Property 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1we have
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 13 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 14 / 41
. . . . . .
Socratic proof
I The definite integral ofvelocity measuresdisplacement (netdistance)
I The derivative ofdisplacement is velocity
I So we can computedisplacement with thedefinite integral or theantiderivative of velocity
I But any function can be avelocity function, so . . .
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 15 / 41
. . . . . .
Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a,b] and f = F′ for another function F, then∫ b
af(x)dx = F(b)− F(a).
NoteIn Section 5.3, this theorem is called “The Evaluation Theorem”.Nobody else in the world calls it that.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
. . . . . .
Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a,b] and f = F′ for another function F, then∫ b
af(x)dx = F(b)− F(a).
NoteIn Section 5.3, this theorem is called “The Evaluation Theorem”.Nobody else in the world calls it that.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 16 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
. . . . . .
Proving the Second FTC
I Divide up [a,b] into n pieces of equal width ∆x =b− an
as usual.
I For each i, F is continuous on [xi−1, xi] and differentiable on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
Orf(ci)∆x = F(xi)− F(xi−1)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 17 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof continued
I We have for each i
f(ci)∆x = F(xi)− F(xi−1)
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 18 / 41
. . . . . .
Proof Completed
I We have shown for each n,
Sn = F(b)− F(a)
Which does not depend on n.
I So in the limit∫ b
af(x)dx = lim
n→∞Sn = lim
n→∞(F(b)− F(a)) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
. . . . . .
Proof Completed
I We have shown for each n,
Sn = F(b)− F(a)
Which does not depend on n.I So in the limit∫ b
af(x)dx = lim
n→∞Sn = lim
n→∞(F(b)− F(a)) = F(b)− F(a)
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 19 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14
.
Here we use the notation F(x)|ba or [F(x)]ba to mean F(b)− F(a).
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14 .
Here we use the notation F(x)|ba or [F(x)]ba to mean F(b)− F(a).
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area between y = x3 and the x-axis, between x = 0 and x = 1.
Solution
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14 .
Here we use the notation F(x)|ba or [F(x)]ba to mean F(b)− F(a).
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 20 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
...−1
..1
..
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
...−1
..1
..
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
. . . . . .
Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x2 and the line y = 1.
...−1
..1
..
1
Solution
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 21 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 22 / 41
. . . . . .
Example
Estimate∫ 1
0
41+ x2
dx using the midpoint rule and four divisions.
SolutionDividing up [0,1] into 4 pieces gives
x0 = 0, x1 =14, x2 =
24, x3 =
34, x4 =
44
So the midpoint rule gives
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=
14
(4
65/64+
473/64
+4
89/64+
4113/64
)=
150,166,78447,720,465
≈ 3.1468
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 23 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10
= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solution
∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10= 4 (arctan 1− arctan 0)
= 4(π4− 0
)= π
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 24 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx
= ln x|21
= ln 2− ln 1= ln 2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 25 / 41
. . . . . .
Estimating an integral with inequalities
Example
Estimate∫ 2
1
1xdx using Property 8.
SolutionSince
1 ≤ x ≤ 2 =⇒ 12≤ 1
x≤ 1
1we have
12· (2− 1) ≤
∫ 2
1
1xdx ≤ 1 · (2− 1)
or12≤
∫ 2
1
1xdx ≤ 1
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 26 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx
= ln x|21
= ln 2− ln 1= ln 2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1= ln 2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1
= ln 2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
. . . . . .
Computing an integral we estimated before
Example
Evaluate∫ 2
1
1xdx.
Solution
∫ 2
1
1xdx = ln x|21
= ln 2− ln 1= ln 2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 27 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 28 / 41
. . . . . .
The Integral as Total Change
Another way to state this theorem is:∫ b
aF′(x)dx = F(b)− F(a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
. . . . . .
The Integral as Total Change
Another way to state this theorem is:∫ b
aF′(x)dx = F(b)− F(a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf v(t) represents the velocity of a particle moving rectilinearly, then∫ t1
t0v(t)dt = s(t1)− s(t0).
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
. . . . . .
The Integral as Total Change
Another way to state this theorem is:∫ b
aF′(x)dx = F(b)− F(a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf MC(x) represents the marginal cost of making x units of a product,then
C(x) = C(0) +∫ x
0MC(q)dq.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
. . . . . .
The Integral as Total Change
Another way to state this theorem is:∫ b
aF′(x)dx = F(b)− F(a),
or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:
TheoremIf ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is
m(x) =∫ x
0ρ(s)ds.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 29 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 30 / 41
. . . . . .
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫
f(x)dx
for any function whose derivative is f(x).
Thus∫x2 dx = 1
3x3 + C.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
. . . . . .
A new notation for antiderivatives
To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫
f(x)dx
for any function whose derivative is f(x). Thus∫x2 dx = 1
3x3 + C.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 31 / 41
. . . . . .
My first table of integrals..
∫[f(x) + g(x)] dx =
∫f(x)dx+
∫g(x)dx∫
xn dx =xn+1
n+ 1+ C (n ̸= −1)∫
ex dx = ex + C∫sin x dx = − cos x+ C∫cos x dx = sin x+ C∫sec2 x dx = tan x+ C∫
sec x tan x dx = sec x+ C∫1
1+ x2dx = arctan x+ C
∫cf(x)dx = c
∫f(x)dx∫
1xdx = ln |x|+ C∫
ax dx =ax
ln a+ C∫
csc2 x dx = − cot x+ C∫csc x cot x dx = − csc x+ C∫
1√1− x2
dx = arcsin x+ C
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 32 / 41
. . . . . .
Outline
Last time: The Definite IntegralThe definite integral as a limitProperties of the integralComparison Properties of the Integral
Evaluating Definite IntegralsExamples
The Integral as Total Change
Indefinite IntegralsMy first table of integrals
Computing Area with integrals
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 33 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.
SolutionThe answer is ∫ 4
1ex dx = ex|41 = e4 − e.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.
SolutionThe answer is ∫ 4
1ex dx = ex|41 = e4 − e.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 34 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solution
I The answer is∫ 1
0arcsin x dx, but we
do not know an antiderivative forarcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solution
I The answer is∫ 1
0arcsin x dx, but we
do not know an antiderivative forarcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solution
I The answer is∫ 1
0arcsin x dx, but we
do not know an antiderivative forarcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy
=π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solution
I The answer is∫ 1
0arcsin x dx, but we
do not know an antiderivative forarcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20
=π
2−1
..x
.
y
..1
..
π/2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
. . . . . .
Computing Area with integrals
Example
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solution
I The answer is∫ 1
0arcsin x dx, but we
do not know an antiderivative forarcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1 ..
x.
y
..1
..
π/2
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 35 / 41
. . . . . .
Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis, andthe vertical lines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2) dx. Notice the integrand is positive on [0,1)
and (2,3], and negative on (1,2). If we want the area of the region, wehave to do
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx+
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10−
[13x
3 − 32x
2 + 2x]21+
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56=
116.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
. . . . . .
Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis, andthe vertical lines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2) dx.
Notice the integrand is positive on [0,1)
and (2,3], and negative on (1,2). If we want the area of the region, wehave to do
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx+
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10−
[13x
3 − 32x
2 + 2x]21+
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56=
116.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 36 / 41
. . . . . .
Graph
.. x.
y
..1
..2
..3
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 37 / 41
. . . . . .
Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis, andthe vertical lines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2) dx. Notice the integrand is positive on [0,1)
and (2,3], and negative on (1,2).
If we want the area of the region, wehave to do
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx+
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10−
[13x
3 − 32x
2 + 2x]21+
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56=
116.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
. . . . . .
Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis, andthe vertical lines x = 0 and x = 3.
Solution
Consider∫ 3
0(x− 1)(x− 2) dx. Notice the integrand is positive on [0,1)
and (2,3], and negative on (1,2). If we want the area of the region, wehave to do
A =
∫ 1
0(x− 1)(x− 2)dx−
∫ 2
1(x− 1)(x− 2)dx+
∫ 3
2(x− 1)(x− 2)dx
=[13x
3 − 32x
2 + 2x]10−
[13x
3 − 32x
2 + 2x]21+
[13x
3 − 32x
2 + 2x]32
=56−
(−16
)+
56=
116.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 38 / 41
. . . . . .
Interpretation of “negative area" in motion
There is an analog in rectlinear motion:
I∫ t1
t0v(t)dt is net distance traveled.
I∫ t1
t0|v(t)|dt is total distance traveled.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 39 / 41
. . . . . .
What about the constant?
I It seems we forgot about the +C when we say for instance∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14− 0 =
14
I But notice[x4
4+ C
]10=
(14+ C
)− (0+ C) =
14+ C− C =
14
no matter what C is.I So in antidifferentiation for definite integrals, the constant is
immaterial.
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 40 / 41
. . . . . .
Summary
I The second Fundamental Theorem of Calculus:∫ b
af(x)dx = F(b)− F(a)
where F′ = f.I Definite integrals represent net change of a function over an
interval.I We write antiderivatives as indefinite integrals
∫f(x)dx
V63.0121.041, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 6, 2010 41 / 41