Visual Aids for Section 6.2Constructing Antiderivatives Analytically
MTH 141
University of Rhode Island
MTH 141 (URI) Section 6.2
Our outcomes
Definitions: indefinite integral, F |ba
Explain why general antiderivatives include “+C” in theirexpressions, and apply “+C” appropriately.
Recall and apply antiderivative rules (eg., for sums/differencesand constant multiples) and antiderivatives for f (x) = k where k isa constant, for f (x) = xn where n 6= −1, for f (x) = 1/x , forf (x) = ex , for f (x) = sin x or f (x) = cos x .
Appropriately use the vertical-bar notation when evaluating anintegral by applying the Fundamental Theorem of Calculus.
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
A puzzle
What is an antiderivative for1
(1− x)2 ?
ddx
(1
1− x
)=
1(1− x)2
So1
(1− x)2 has a family of antiderivatives, right?
11− x
+ anyconstant
Now what’sddx
(x
1− x
)?
ddx
(x
1− x
)=
1(1− x)2
Weird! What’s going on? And how would we ever come up with all thepossible antiderivatives?
MTH 141 (URI) Section 6.2
The family of antiderivatives of a function f
Two key facts:
If F ′(x) = 0 everywhere on an interval, then F (x) = C on thisinterval, where C a single constant.
If G and H are both antiderivatives of f on the same interval, thenG(x) = H(x) + C, where C is some constant.
A key consequence of the second fact is that the generalantiderivative of a function will always include a “+C”.∫
2x dx = x2+C;
∫cos x dx = sin(x)+C.
Notation:∫... dx means an antiderivative (also known as an
indefinite integral), and +C indicates that any constant may beadded, still producing an antiderivative.
MTH 141 (URI) Section 6.2
The family of antiderivatives of a function f
Two key facts:
If F ′(x) = 0 everywhere on an interval, then F (x) = C on thisinterval, where C a single constant.
If G and H are both antiderivatives of f on the same interval, thenG(x) = H(x) + C, where C is some constant.
A key consequence of the second fact is that the generalantiderivative of a function will always include a “+C”.∫
2x dx = x2+C;
∫cos x dx = sin(x)+C.
Notation:∫... dx means an antiderivative (also known as an
indefinite integral), and +C indicates that any constant may beadded, still producing an antiderivative.
MTH 141 (URI) Section 6.2
A puzzle
ddx
(1
1− x
)=
1(1− x)2 and
ddx
(x
1− x
)=
1(1− x)2
Weird...What’s going on?
How are1
1− xand
x1− x
both antiderivatives of1
(1− x)2 ?
...Becausex
1− x=
11− x
− 1!
MTH 141 (URI) Section 6.2
A puzzle
ddx
(1
1− x
)=
1(1− x)2 and
ddx
(x
1− x
)=
1(1− x)2
Weird...What’s going on?
How are1
1− xand
x1− x
both antiderivatives of1
(1− x)2 ?
...Becausex
1− x=
11− x
− 1!
MTH 141 (URI) Section 6.2
The family of antiderivatives of a function f
Two key facts:
If F ′(x) = 0 everywhere on an interval, then F (x) = C on thisinterval, where C a single constant.
If G and H are both antiderivatives of f on the same interval, thenG(x) = H(x) + C, where C is some constant.
A key consequence of the second fact is that the generalantiderivative of a function will always include a “+C”.∫
2x dx = x2+C;
∫cos x dx = sin(x)+C.
Notation:∫... dx means an antiderivative (also known as an
indefinite integral), and +C indicates that any constant may beadded, still producing an antiderivative.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .
∫x7 dx =
18
x8
+ C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =
18
x8
+ C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8
+ C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8 + C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8 + C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8 + C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8 + C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C,
we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
Let’s try it out
Evaluate∫
x7 dx .∫
x7 dx =18
x8 + C
Find a general antiderivative for the
function1x
shown at the right.
Instead of∫ 1
x dx = ln x + C, we
write∫
1x
dx = ln |x |+ C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx =
kx + C.
If n is any real number otherthan −1, then∫
xn dx =
1n + 1
xn+1 + C.
∫1x
dx =
ln |x |+ C.
∫ex dx =
ex + C.
∫cos x dx =
sin(x) + C.
∫sin x dx =
− cos(x) + C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx = kx + C.
If n is any real number otherthan −1, then∫
xn dx =
1n + 1
xn+1 + C.
∫1x
dx =
ln |x |+ C.
∫ex dx =
ex + C.
∫cos x dx =
sin(x) + C.
∫sin x dx =
− cos(x) + C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx = kx + C.
If n is any real number otherthan −1, then∫
xn dx =1
n + 1xn+1 + C.
∫1x
dx =
ln |x |+ C.
∫ex dx =
ex + C.
∫cos x dx =
sin(x) + C.
∫sin x dx =
− cos(x) + C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx = kx + C.
If n is any real number otherthan −1, then∫
xn dx =1
n + 1xn+1 + C.
∫1x
dx = ln |x |+ C.∫ex dx =
ex + C.
∫cos x dx =
sin(x) + C.
∫sin x dx =
− cos(x) + C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx = kx + C.
If n is any real number otherthan −1, then∫
xn dx =1
n + 1xn+1 + C.
∫1x
dx = ln |x |+ C.∫ex dx = ex + C.∫cos x dx =
sin(x) + C.
∫sin x dx =
− cos(x) + C.
MTH 141 (URI) Section 6.2
A few antiderivatives to know by heart
If k is a constant, then∫k dx = kx + C.
If n is any real number otherthan −1, then∫
xn dx =1
n + 1xn+1 + C.
∫1x
dx = ln |x |+ C.∫ex dx = ex + C.∫cos x dx = sin(x) + C.∫sin x dx = − cos(x) + C.
MTH 141 (URI) Section 6.2
Properties of antiderivativesSums and constant multiples
∫(f (x) + g(x)) dx =
∫f (x) dx +
∫g(x) dx
∫(f (x)− g(x)) dx =
∫f (x) dx −
∫g(x) dx
∫cf (x) dx =
c∫
f (x) dx
MTH 141 (URI) Section 6.2
Properties of antiderivativesSums and constant multiples
∫(f (x) + g(x)) dx =
∫f (x) dx +
∫g(x) dx
∫(f (x)− g(x)) dx =
∫f (x) dx −
∫g(x) dx
∫cf (x) dx =
c∫
f (x) dx
MTH 141 (URI) Section 6.2
Properties of antiderivativesSums and constant multiples
∫(f (x) + g(x)) dx =
∫f (x) dx +
∫g(x) dx
∫(f (x)− g(x)) dx =
∫f (x) dx −
∫g(x) dx
∫cf (x) dx = c
∫f (x) dx
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
2 Find the general antiderivative in each of the following:
(a)∫
cos x dx (b)∫
sin x dx (c)∫
ex dx
(d)∫
(x3 + x2 + x +√
x + 7 +1x2 ) dx
(e)∫
x + 1x
dx (f)∫(sinh(t) + t2(t3 − 5)) dt
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =
12
sin2x
+C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =
12
sin2x
+C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x
+C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).
We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
1 Find an antiderivative of cos2x which passes through the point(π,37).
We start by guessing an antiderivative:∫
cos2x dx =12sin2x +C
This is the general antiderivative (or indefinite integral). We want aspecific antiderivative, the one passing through (π,37).We’ll get the right answer if we can figure out what C should equal.
Plugging in π and 37, we get12sin(2 · π) + C = 37
0 + C = 37
Knowing what C should be now, we find that the desired antiderivativeis
12sin2x + 37.
MTH 141 (URI) Section 6.2
Exercises
2 Find the general antiderivative in each of the following:
(a)∫
cos x dx (b)∫
sin x dx (c)∫
ex dx
= sin x +C = − cos x +C = ex +C
(d)∫(x3 + x2 + x +
√x + 7 +
1x2 ) dx
= 14x4 + 1
3x3 + 12x2 + 1
3/2x3/2 + 7x + 1−1x−1 + C
(e)∫
x + 1x
dx (f)∫(sinh(t) + t2(t3 − 5)) dt
MTH 141 (URI) Section 6.2
Exercises
2 Find the general antiderivative in each of the following:
(a)∫
cos x dx (b)∫
sin x dx (c)∫
ex dx
= sin x +C = − cos x +C = ex +C
(d)∫(x3 + x2 + x +
√x + 7 +
1x2 ) dx
= 14x4 + 1
3x3 + 12x2 + 1
3/2x3/2 + 7x + 1−1x−1 + C
(e)∫
x + 1x
dx (f)∫(sinh(t) + t2(t3 − 5)) dt
MTH 141 (URI) Section 6.2
Exercises
2 Find the general antiderivative in each of the following:
(a)∫
cos x dx (b)∫
sin x dx (c)∫
ex dx
= sin x +C = − cos x +C = ex +C
(d)∫(x3 + x2 + x +
√x + 7 +
1x2 ) dx
= 14x4 + 1
3x3 + 12x2 + 1
3/2x3/2 + 7x + 1−1x−1 + C
(e)∫
x + 1x
dx (f)∫(sinh(t) + t2(t3 − 5)) dt
MTH 141 (URI) Section 6.2
Exercises
2 Find the general antiderivative in each of the following:
(a)∫
cos x dx (b)∫
sin x dx (c)∫
ex dx
= sin x +C = − cos x +C = ex +C
(d)∫(x3 + x2 + x +
√x + 7 +
1x2 ) dx
= 14x4 + 1
3x3 + 12x2 + 2
3x3/2 + 7x − 1x + C
(e)∫
x + 1x
dx (f)∫(sinh(t) + t2(t3 − 5)) dt
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx
=
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx
= x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt =
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt =
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt =
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Exercises
(e)∫
x + 1x
dx =
∫ (1 +
1x
)dx = x + ln |x |+ C
(f)∫(sinh(t) + t2(t3 − 5)) dt =
∫sinh(t) dt +
∫t2(t3 − 5)) dt
=
∫sinh(t) dt +
∫(t5 − 5t2)dt
=
∫sinh(t) dt +
∫t5 dt − 5
∫t2 dt
= cosh(t) +16
t6 − 53
t3 + C
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx
= F (x)∣∣∣∣ba
= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41=
13
43 − 13
13 =64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41=
13
43 − 13
13 =64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41=
13
43 − 13
13 =64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41
=13
43 − 13
13 =64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41=
13
43 − 13
13
=64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example: ∫ 4
1x2 dx =
13
x3∣∣∣∣41=
13
43 − 13
13 =64− 1
3= 21.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example:∫ π/4
0
1cos2 t
dt =
tan t∣∣∣∣π/4
0= tan(π/4)− tan0 = 1− 0 = 1.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example:∫ π/4
0
1cos2 t
dt = tan t∣∣∣∣π/4
0
= tan(π/4)− tan0 = 1− 0 = 1.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example:∫ π/4
0
1cos2 t
dt = tan t∣∣∣∣π/4
0= tan(π/4)− tan0
= 1− 0 = 1.
MTH 141 (URI) Section 6.2
Back to the Fundamental Theorem of Calculus
Suppose F is an antiderivative of f . Then∫ b
af (x) dx = F (x)
∣∣∣∣ba= F (b)− F (a)
Notation: When simplifying∫ b
a f (x) dx using the FTC, we use∫ b
a ... dxwhen we have not yet taken an antiderivative of the function f (x). Wewrite the vertical bar following F (x) after we’ve taken an antiderivative.
Example:∫ π/4
0
1cos2 t
dt = tan t∣∣∣∣π/4
0= tan(π/4)− tan0 = 1− 0 = 1.
MTH 141 (URI) Section 6.2
Our outcomes
Definitions: indefinite integral, F |ba
Explain why general antiderivatives include “+C” in theirexpressions, and apply “+C” appropriately.
Recall and apply antiderivative rules (eg., for sums/differencesand constant multiples) and antiderivatives for f (x) = k where k isa constant, for f (x) = xn where n 6= −1, for f (x) = 1/x , forf (x) = ex , for f (x) = sin x or f (x) = cos x .
Appropriately use the vertical-bar notation when evaluating anintegral by applying the Fundamental Theorem of Calculus.
MTH 141 (URI) Section 6.2
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