mat01a1: antiderivatives · are also antiderivatives of f(x) = 2xbecause g0(x) = 2x and h0(x) = 2x....
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MAT01A1: Antiderivatives
Dr Craig
Week: 11 May 2020
Warm up
Calculate the limit: limx→−4
sin(πx)
x2 − 16
Attempt this on your own before looking at
the solution on the next slide.
limx→−4
sin(πx)
x2 − 16We have lim
x→−4sin(πx) = sin(−4π) = 0 and
limx→−4
(x2 − 16) = (−4)2 − 16 = 0, so we
have a 00 indeterminate form.
limx→−4
sin(πx)
x2 − 16
(0
0
)L’H= lim
x→−4
π cos(πx)
2x
=π cos(−4π)
2(−4)=
π
−8≈ −0.393
limx→−4
sin(πx)
x2 − 16We have lim
x→−4sin(πx) = sin(−4π) = 0 and
limx→−4
(x2 − 16) = (−4)2 − 16 = 0, so we
have a 00 indeterminate form.
limx→−4
sin(πx)
x2 − 16
(0
0
)L’H= lim
x→−4
π cos(πx)
2x
=π cos(−4π)
2(−4)=
π
−8≈ −0.393
Warm up
Calculate the limit: limx→−4
sin(πx)
x2 − 16
An ∞−∞ indeterminate form example
limx→∞
(lnx− x)
Attempt this before looking at the solution
on the next slide.
Hint: take out a “common factor”.
An ∞−∞ indeterminate form example
limx→∞
(lnx− x)
Attempt this before looking at the solution
on the next slide.
Hint: take out a “common factor”.
limx→∞
(lnx− x)
Solution:
limx→∞
(lnx− x) = limx→∞
x
(lnx
x− 1
)We want to see what happens to lnx
x as
x→∞.
limx→∞
lnx
x
(∞∞
)L’H= lim
x→∞
1/x
1= lim
x→∞
1
x= 0
limx→∞
(lnx− x)Solution:
limx→∞
(lnx− x) = limx→∞
x
(lnx
x− 1
)
We want to see what happens to lnxx as
x→∞.
limx→∞
lnx
x
(∞∞
)L’H= lim
x→∞
1/x
1= lim
x→∞
1
x= 0
limx→∞
(lnx− x)Solution:
limx→∞
(lnx− x) = limx→∞
x
(lnx
x− 1
)We want to see what happens to lnx
x as
x→∞.
limx→∞
lnx
x
(∞∞
)L’H= lim
x→∞
1/x
1= lim
x→∞
1
x= 0
limx→∞
(lnx− x)Solution:
limx→∞
(lnx− x) = limx→∞
x
(lnx
x− 1
)We want to see what happens to lnx
x as
x→∞.
limx→∞
lnx
x
(∞∞
)L’H= lim
x→∞
1/x
1= lim
x→∞
1
x= 0
limx→∞
(lnx− x)
Solution continued. . . We have
limx→∞
lnx
x= 0 so
limx→∞
(lnx
x− 1
)= −1.
Therefore
limx→∞
x
(lnx
x− 1
)= −∞
limx→∞
(lnx− x)
Solution continued. . . We have
limx→∞
lnx
x= 0 so
limx→∞
(lnx
x− 1
)= −1.
Therefore
limx→∞
x
(lnx
x− 1
)= −∞
After that brief warm-up, we move on to the
topic of Ch 4.9: antiderivatives. You can
probably already guess what is contained in
this section.
Chapter 3 was all about how to go from a
function f to its derivative f ′. This section
is about going in the opposite direction.
That is, if we are given some function f ,
what is the function F such that F ′ = f .
Antiderivatives
A function F is an antiderivative of f on
an interval I if
F ′(x) = f (x)
for all x ∈ I .
Example: F (x) = x2 is an antiderivative of
f (x) = 2x.
Question: are antiderivatives unique?
Antiderivatives
A function F is an antiderivative of f on
an interval I if
F ′(x) = f (x)
for all x ∈ I .
Example: F (x) = x2 is an antiderivative of
f (x) = 2x.
Question: are antiderivatives unique?
Antiderivatives
A function F is an antiderivative of f on
an interval I if
F ′(x) = f (x)
for all x ∈ I .
Example: F (x) = x2 is an antiderivative of
f (x) = 2x.
Question: are antiderivatives unique?
Answer: No, antiderivatives are not unique.
Consider the previous example of f (x) = 2x
and F (x) = x2. The functions
G(x) = x2 + 4 or H(x) = x2 + 7
are also antiderivatives of f (x) = 2x because
G′(x) = 2x and H ′(x) = 2x
Answer: No, antiderivatives are not unique.
Consider the previous example of f (x) = 2x
and F (x) = x2.
The functions
G(x) = x2 + 4 or H(x) = x2 + 7
are also antiderivatives of f (x) = 2x because
G′(x) = 2x and H ′(x) = 2x
Answer: No, antiderivatives are not unique.
Consider the previous example of f (x) = 2x
and F (x) = x2. The functions
G(x) = x2 + 4 or H(x) = x2 + 7
are also antiderivatives of f (x) = 2x because
G′(x) = 2x and H ′(x) = 2x
Theorem: If F is an antiderivative of f
on an interval I , then the most general
antiderivative of f on I is F (x) + C
where C is an arbitrary constant.
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3
+ x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2
+ 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x
+ C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3
+3
2x2 + 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2
+ 4x + C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x
+ C
Examples of general antiderivatives
Find the general antiderivative of
f (x) = 3x2 + 2x + 6
Answer: F (x) = x3 + x2 + 6x + C
Harder: find the general antiderivative of
g(x) = 7x2 + 3x + 4
Answer: G(x) =7
3x3 +
3
2x2 + 4x + C
More examples:
Find the general antiderivative of
f (x) = sinx
Answer: F (x) = − cosx + C
More examples:
Find the general antiderivative of
f (x) = sinx
Answer: F (x) = − cosx + C
Function Particular antiderivative
cf (x) cF (x)
f (x) + g(x) F (x) +G(x)
xn (n 6= −1) 1n+1x
n+1
1
x`n|x|
ex ex
Function Particular antiderivative
cosx sinx
sinx − cosx
sec2 x tanx
secx tanx secx
Function Particular antiderivative1√
1− x2arcsinx
1
1 + x2arctanx
coshx sinhx
sinhx coshx
In the examples on this slide, we are not
looking for the most general antiderivative,
but for the antiderivative that satisfies some
additional conditions.
Examples:
1. Find f if f ′′(x) = 12x2 + 6x− 4,
f (0) = 4 and f (1) = 1.
2. Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.):
Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4.
Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3.
Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′′(x) = 12x2 + 6x− 4, f (0) = 4
and f (1) = 1.
Solution (1.): Using the reverse of the power
rule we get f ′(x) = 4x3 + 3x2 − 4x + C.
This gives f (x)=x4 + x3 − 2x2 + Cx +D.
Now we substitute x = 0 and get
4 = 04 + 03 − 2(02) + C(0) +D,
so D = 4. Now substitute x = 1 and get
1 = 14 + 13 − 2(12) + C(1) + 4,
so C = −3. Our final function is
f (x) = x4 + x3 − 2x2 − 3x + 4.
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.):
Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C. We
substitute x = 0 and get
−2 = e0+20 arctan 0+C =⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.): Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C. We
substitute x = 0 and get
−2 = e0+20 arctan 0+C =⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.): Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C.
We
substitute x = 0 and get
−2 = e0+20 arctan 0+C =⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.): Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C. We
substitute x = 0 and get
−2 = e0+20 arctan 0+C
=⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.): Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C. We
substitute x = 0 and get
−2 = e0+20 arctan 0+C =⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
Find f if f ′(x) = ex + 20(1 + x2)−1 and
f (0) = −2.
Solution (2.): Recall that
d
dx(arctanx) =
1
1 + x2.
Therefore f (x) = ex + 20 arctanx + C. We
substitute x = 0 and get
−2 = e0+20 arctan 0+C =⇒ C = −3.
Therefore f (x) = ex + 20 arctan(x)− 3
REMEMBER THE CHAIN RULE!
Find the most general antiderivative of:
f (x) = cosx. sec2(sinx)
We will only learn a formal method for this in
Ch 5.5 but it is worth introducing the idea
already now. Notice that cosx is the
derivative of sinx which is the inner function
of the composite function sec2(sinx).
Looking at this (and thinking hard) we get:
F (x) = tan(sinx) + C
REMEMBER THE CHAIN RULE!
Find the most general antiderivative of:
f (x) = cosx. sec2(sinx)
We will only learn a formal method for this in
Ch 5.5 but it is worth introducing the idea
already now.
Notice that cosx is the
derivative of sinx which is the inner function
of the composite function sec2(sinx).
Looking at this (and thinking hard) we get:
F (x) = tan(sinx) + C
REMEMBER THE CHAIN RULE!
Find the most general antiderivative of:
f (x) = cosx. sec2(sinx)
We will only learn a formal method for this in
Ch 5.5 but it is worth introducing the idea
already now. Notice that cosx is the
derivative of sinx which is the inner function
of the composite function sec2(sinx).
Looking at this (and thinking hard) we get:
F (x) = tan(sinx) + C
REMEMBER THE CHAIN RULE!
Find the most general antiderivative of:
f (x) = cosx. sec2(sinx)
We will only learn a formal method for this in
Ch 5.5 but it is worth introducing the idea
already now. Notice that cosx is the
derivative of sinx which is the inner function
of the composite function sec2(sinx).
Looking at this (and thinking hard) we get:
F (x) = tan(sinx) + C
Below is another example of this type.
Find the most general antiderivative of
f (x) = 3x2ex3
Notice here that ddx(x
3) = 3x2. From this we
get:
F (x) = ex3+ C
As we mentioned, we will introduce a formal
technique for dealing with antiderivatives of
this type a bit later on.
Below is another example of this type.
Find the most general antiderivative of
f (x) = 3x2ex3
Notice here that ddx(x
3) = 3x2. From this we
get:
F (x) = ex3+ C
As we mentioned, we will introduce a formal
technique for dealing with antiderivatives of
this type a bit later on.
Below is another example of this type.
Find the most general antiderivative of
f (x) = 3x2ex3
Notice here that ddx(x
3) = 3x2.
From this we
get:
F (x) = ex3+ C
As we mentioned, we will introduce a formal
technique for dealing with antiderivatives of
this type a bit later on.
Below is another example of this type.
Find the most general antiderivative of
f (x) = 3x2ex3
Notice here that ddx(x
3) = 3x2. From this we
get:
F (x) = ex3+ C
As we mentioned, we will introduce a formal
technique for dealing with antiderivatives of
this type a bit later on.
Below is another example of this type.
Find the most general antiderivative of
f (x) = 3x2ex3
Notice here that ddx(x
3) = 3x2. From this we
get:
F (x) = ex3+ C
As we mentioned, we will introduce a formal
technique for dealing with antiderivatives of
this type a bit later on.
Prescribed tut problems:
Complete the following exercises from the
8th edition (Ch 4.9):
1, 3, 5, 11, 13, 15, 17, 25, 29, 37, 41
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