Visual Aids for Section 6.2 Constructing Antiderivatives ...barrus/mth141/sp20/Slides62.pdfThe...

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.


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?


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.


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!


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.


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.


Let’s try it out

Evaluate∫

x7 dx .∫

x7 dx =

18

x8

+ C


function1x

shown at the right.

Instead of∫ 1

x dx = ln x + C,

we

write∫

1x

dx = ln |x |+ C.


Let’s try it out

Evaluate∫

x7 dx .∫

x7 dx =18

x8

+ C


function1x

shown at the right.

Instead of∫ 1

x dx = ln x + C,

we

write∫

1x

dx = ln |x |+ C.


Let’s try it out

Evaluate∫

x7 dx .∫

x7 dx =18

x8 + C


function1x

shown at the right.

Instead of∫ 1

x dx = ln x + C,

we

write∫

1x

dx = ln |x |+ C.


Let’s try it out

Evaluate∫

x7 dx .∫

x7 dx =18

x8 + C


function1x

shown at the right.

Instead of∫ 1

x dx = ln x + C, we

write∫

1x

dx = ln |x |+ C.


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.



If k is a constant, then∫k dx = kx + C.


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.





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.





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.





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.





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.


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


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


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


Exercises


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.


Exercises



cos2x dx =12sin2x

+C



0 + C = 37


12sin2x + 37.


Exercises



cos2x dx =12sin2x +C



0 + C = 37


12sin2x + 37.


Exercises




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.


0 + C = 37


12sin2x + 37.


Exercises






0 + C = 37


12sin2x + 37.


Exercises


(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


Exercises


(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


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


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


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


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


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.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example: ∫ 4

1x2 dx =

13

x3∣∣∣∣41=

13

43 − 13

13 =64− 1

3= 21.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example: ∫ 4

1x2 dx =

13

x3∣∣∣∣41

=13

43 − 13

13 =64− 1

3= 21.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example: ∫ 4

1x2 dx =

13

x3∣∣∣∣41=

13

43 − 13

13

=64− 1

3= 21.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example: ∫ 4

1x2 dx =

13

x3∣∣∣∣41=

13

43 − 13

13 =64− 1

3= 21.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example:∫ π/4

0

1cos2 t

dt =

tan t∣∣∣∣π/4

0= tan(π/4)− tan0 = 1− 0 = 1.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example:∫ π/4

0

1cos2 t

dt = tan t∣∣∣∣π/4

0

= tan(π/4)− tan0 = 1− 0 = 1.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example:∫ π/4

0

1cos2 t


0= tan(π/4)− tan0

= 1− 0 = 1.




af (x) dx = F (x)

∣∣∣∣ba= F (b)− F (a)




Example:∫ π/4

0

1cos2 t


0= tan(π/4)− tan0 = 1− 0 = 1.


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.


Visual Aids for Section 6.2 Constructing Antiderivatives ...barrus/mth141/sp20/Slides62.pdfThe...

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