The FTC Part 2, Total Change/Area & U-Sub

Question from Test 1

Liquid drains into a tank at the rate 21e-3t units per minute. If the tank starts empty and can hold 6 units, at what time will it overflow?

A. log(7)/3 B. (1/3)log(13/7) C. 3 log (13/7) D. 3log(7) E. Never

Question from Test 1

Liquid drains into a tank at the rate 21e-3t units per minute. If the tank starts empty and can hold 6 units, at what time will it overflow?

A. log(7)/3

The Total Change Theorem

′F x( )a

b

∫ dx = F b( ) − F a( )

The integral of a rate of change is the total change from a to b. (displacement)

(still from last weeks notes)

The Total Change Theorem Ex: Given

Find the displacement and total distance traveled from time 1 to time 6. Displacement: (negative area takes away from positive)

Total Distance: (all area counted positive)

v(t) =t2 −4t+ 3

t 2 −4t+ 3( )dt1

6

∫

− t 2 − 4t + 3( )dt1

3

∫ + t 2 − 4t + 3( )dt3

6

∫

Total Area Find the area of the region bounded by

the x-axis, y-axis and y = 2 – 2x. First find the bounds by setting 2 – 2x = 0 and by substitution 0 in for x

€

2 − 2x( )dx0

1∫

Total Area Ex. Find the area of the region

bounded by the y-axis and the curve

€

x = 2y2 + 3y4 − 2y6

€

2y2 + 3y4 − 2y6( )0

2∫ dy

Fundamental Theorem of Calculus (Part 1)(Chain Rule)

If f is continuous on [a, b], then the function defined by

is continuous on [a, b] and differentiable on (a, b) and

g(x) = f(t)dta

u(x)

∫ a≤x≤b

g '(x) = f(u(x))u'(x)

Fundamental Theorem of Calculus (Part 1)

d

drv eu2 v −1 dv

1

r3

∫ 3r2( )r3 eu2r3 −1

d

drv eu2 v −1 dv

1

r3

∫


If f is continuous on [a, b], then :

Where F is any antiderivative of f.

( )

f (t)dta

b

∫ =F(b)−F(a)

F ' = f

Helps us to more easily evaluate Definite Integrals in the same way we calculate the Indefinite!

Example

Example

We have to • find an antiderivative;• evaluate at 3;• evaluate at 2;• subtract the results.

Example

Example

This notation means: evaluate the function at 3 and 2, and subtract the

results.

Example

Example

Don’t need to include “+ C” in our antiderivative, because any antiderivative

will work.

Example

the “C’s” will cancel each other out.

x3dx2

3

∫ =14x4 +C⎡

⎣⎢⎤⎦⎥2

3

=1

434 + C

⎛⎝⎜

⎞⎠⎟

−1

424 + C

⎛⎝⎜

⎞⎠⎟

=1

434 + C −

1

424 − C

=81

4−

16

4=

65

4

Example

Example

Alternate notation

Example

Example

= –1

Example

= –1 = 1

Example

Given: f (x) =

0 x< 0x 0 ≤x≤12 −x 1 < x≤20 x> 2

⎧

⎨⎪⎪

⎩⎪⎪

Write a similar expression for the continuous function:

g(x) = f (t)dt0

x

∫g(x) =

0 x< 0

x2

2 0 ≤x≤1

2x−x2

2−1 1 < x≤2

1 x> 2

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪


If f is continuous on [a, b], then :

Where F is any antiderivative of f. ( )

f (t)dta

b

∫ =F(b)−F(a)

F ' = f

Evaluate:

Multiply out:

2x +1( )2 dx−1

2

∫

= 4x2 + 4x +1( )dx−1

2

∫

Use FTC 2 to Evaluate:

=[4

3x3 + 2x + x

−1

2

=4

3(2)3 + 2(2)2 + 2 − (

4

3(−1)3 + 2(−1)2 −1)

=21

What if instead?

It would be tedious to use the same multiplication strategy!

There is a better way!

We’ll use the chain rule (backwards)

2x +1( )10 dx∫

Chain Rule for Derivatives:

d

dxf g x( )( )⎡⎣ ⎤⎦= f ' g(x)( )g'(x)

Chain Rule backwards for Integration:

f ' g(x)( )g'(x)dx=∫ f g x( )( ) +C

Look for:

f ' g(x)( ) g'(x)dx=∫ f g x( )( ) +C

Back to Our Example

Let

2x +1( )2 dx−1

2

∫

2x +1( )2dx

−1

2

∫u =2x+1du =2dx

=1

22x +1( )

22dx

−1

2

∫2x +1( )2 dx−1

2

∫

Our Example as anIndefinite Integral

With

AND Without worrying about the bounds for now:

Back to x (Indefinite):

2x +1( )2 dx∫

u =2x+1du =2dx

1

22x +1( )2 2dx∫

=1

62x +1( )

3+ C

=1

2u2du∫ =

1

6u3 + C

The same substitution holds for the higher power!

With

Back to x (Indefinite):

2x +1( )10 dx∫

u =2x+1du =2dx

1

22x +1( )10 2dx∫

=1

222x +1( )

11+ C

=1

2u10du∫ =

1

22u11 + C

Our Original Exampleof a Definite Integral:

To make the substitution complete for a Definite Integral: We make a change of bounds using:

When x = -1, u = 2(-1)+1 = -1 When x = 2, u = 2(2) + 1 = 5

The x-interval [-1,2] is transformed to the u-interval [-1, 5]

2x +1( )2 dx−1

2

∫

u =2x+1

1

22x +1( )2 2dx

−1

2

∫

=16

53 −(−1)3( )

=1

2u2du

−1

5

∫ =1

6u3

−1

5

=21

Substitution Rule for Indefinite Integrals

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

f g(x)( )g'(x)dx= f(u)du∫∫Substitution Rule for Definite Integrals

If g’(x) is continuous on [a,b] and f is continuous on the range of u = g(x), then

f g(x)( )g'(x)dxa

b

∫ = f(u)dug(a)

g(b)

∫

Using the Chain Rule, we know that:

d

dxsin x2( )( ) =2xcos x2( )

Evaluate:

Evaluate:

Looks almost like cos(x2) 2x, which is

the derivative of sin(x2).


d


Evaluate:

We will rearrange the integral to get an exact match:

d



Evaluate:



d


x cos x2( )dx= cos x2( )xdx∫∫

Evaluate:



d


x cos x2( )dx= cos x2( )xdx∫∫=12

cos x2( )2xdx∫

Evaluate:


We put in a 2 so the pattern will match.


d



cos x2( )2xdx∫

Evaluate:


We put in a 2 so the pattern will match.

So we must also put in a 1/2 to keep the

problem the same.


d



cos x2( )2xdx∫

Evaluate:



d



cos x2( )2xdx∫

Evaluate:



d



cos x2( )2xdx∫

=12sin x2( )⎡⎣ ⎤⎦+C

Check Answer:

Check:

Check Answer:

Check:

From the chain rule

Check Answer:

1) Choose u.

Indefinite Integrals by Substitution

1) Choose u.

2) Calculate du.

€

du =du

dxdx


1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)€

du =du

dxdx


1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)

4) Solve the new integral.

€

du =du

dxdx



1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)

4) Solve the new integral.

5) Substitute back in to get x again.

€

du =du

dxdx

ExampleA linear substitution:

Let u = 3x + 2. Then du = 3dx.

e3x+2dx∫ =13

e3x+23dx∫

=1

3eudu∫

=1

3eu + C

=1

3e3x+2 + C

Choosing u Try to choose u to be an inside function.

(Think chain rule.) Try to choose u so that du is in the

problem, except for a constant multiple.

Choosing uFor

u = 3x + 2 was a good choice because

(1) 3x + 2 is inside the exponential.

(2) The derivative is 3, which is only a constant.

PracticeLet u =

du =

PracticeLet u = x2 + 1

du =


du = 2x dx


du = 2x dx

Make this a 2x dx and we’re

set!


du = 2x dx


du = 2x dx

= 1

2ln x2 + 1 + C

PracticeLet u =

du =

PracticeLet u = sin(x)

du =

PracticeLet u = sin(x)

du = cos(x) dx

PracticeLet u =

du = An alternate possibility:

PracticeLet u = cos(x)

du = –sin(x) dxAn alternate possibility:

Practice

Note:

Practice

Note:

What’s the difference?

PracticeNote:


Practice

Note:


Practice

Note:


This is 1!

Practice

Note:


Practice

Note:


That is, the difference is a constant.

In-Class Assignment

Integrate using two different methods:

1st by multiplying out and integrating 2nd by u-substitution

Do you get the same result? (Don’t just assume or claim you do; multiply out your results to show it!)

If you don’t get exactly the same answer, is it a problem? Why or why not?

3x −1( )∫2dx

The FTC Part 2, Total Change/Area & U-Sub

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