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
The integral of a rate of change is the total change from
a to b. (displacement)

b
a
F  x  dx  F b   F a 
(still from last weeks notes)
The Total Change Theorem

Ex: Given
v(t)  t 2  4t  3


Find the displacement and total distance traveled from time 1 to
time 6.
Displacement: (negative area takes away from positive)
6


t 2  4t  3 dt

Total Distance: (all area1 counted positive)
3


6


  t 2  4t  3 dt   t 2  4t  3 dt
1
3

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
 0 2  2xdx
1

Total Area

Ex. Find the area of the region
bounded by the y-axis and the curve
x  2y 2  3y 4  2y 6
 0 2y
2
2

 3y  2y dy
4
6
Fundamental Theorem of Calculus (Part 1)
(Chain Rule)

If f is continuous on [a, b], then the function
defined by
u( x )
g(x) 

f (t)dt a  x  b
a

is continuous on [a, b] and differentiable on (a, b)
and
g'(x)  f (u(x))u '(x)
Fundamental Theorem of Calculus (Part 1)
r3
d
u2 v
v e  1 dv

dr 1
3r r
2
3
u2 r 3
e
1
Fundamental Theorem of Calculus (Part 2)

If f is continuous on [a, b], then :
b
 f (t)dt  F(b)  F(a)
a

Where F is any antiderivative of f.
( 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
Example
Example
Don’t need to include “+ C” in our
antiderivative, because any antiderivative
will work.
Example
3
3
1 4

3
x
dx

x

C
2
 4

2
1
 1

  34  C    2 4  C 
4
 4


1 4
1
3  C  24  C
4
4

81 16 65


4
4
4
the “C’s” will cancel each other out.
Example
Example
Example
Alternate notation
Example
Example
Example
= –1
Example
= –1
=1
Example
Example
Example
x0
0
x
0  x 1

Given: f (x)  2  x 1  x  2

0
x2
Write a similar expression for the continuous
function:
x0
0
x
 2
g(x)   f (t)dt
x

0  x 1
0
2
g(x)  
2
x
2x 
1 1 x  2

2
1
x2

Fundamental Theorem of Calculus (Part 2)

If f is continuous on [a, b], then :
b
 f (t)dt  F(b)  F(a)
a
Where F is any antiderivative of f.
( F'  f )
2
 2x  1 dx
2
Evaluate:
1
2

Multiply out:

 4x
2

 4x  1 dx
1
Use FTC 2 to Evaluate:
2
4 3
 [ x  2x  x 1
3
4 3
4
2
 (2)  2(2)  2  ( (1)3  2(1)2  1)
3
3
 21
 2x  1
10
What if instead?
dx

It would be tedious to use the same
multiplication strategy!

There is a better way!

We’ll use the chain rule (backwards)
Chain Rule for Derivatives:
d
 f g x   f ' g(x)g '(x)
dx

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
2
 2x  1 dx
2
Back to Our Example

Let
 2x  1 dx
2
1
 2x  1  dx
2
u  2x  1
du  2dx
2
1
1
2
1
2
  2x  1 2dx
2 1
2
Our Example as an
Indefinite Integral
 2x  1 dx
2
u  2x  1
du  2dx

With

AND Without worrying about the bounds for now:
1 3
1 2
1
2
2x  1 2dx   u du  6 u  C

2
2

Back to x (Indefinite):
1
3
 2x  1  C
6
The same substitution
holds for the higher power!

With
 2x  1
10
dx
u  2x  1
du  2dx
1 10
1
10
1 11
2x  1 2dx   u du  u  C

2
2
22

Back to x (Indefinite):
1
11

2x  1  C
22
2
Our Original Example
of a Definite Integral:



 2x  1 dx
2
1
To make the substitution complete for a Definite Integral:
We make a change of bounds using:
u  2x  1
 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]
2
5
5
1 3
1 2
1
2
2x  1 2dx   u du  6 u

2 1
2 1
1

1 3
 5  (1)3
6

 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
b
g(b)
 f g(x)g '(x)dx  
a
g(a)
f (u)du
Using the Chain Rule,
we know that:
Evaluate:
   2x cos x 
d
sin x 2
dx
2
Using the Chain Rule,
we know that:
Evaluate:
Looks almost like
cos(x2) 2x, which is
the derivative of
sin(x2).
   2x cos x 
d
sin x 2
dx
2
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
Evaluate:
We will rearrange the integral to get an exact match:
2
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
Evaluate:
We will rearrange the integral to get an exact match:
 
 
2
2
x
cos
x
dx

cos
x
xdx


2
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
Evaluate:
We will rearrange the integral to get an exact match:
 
 
1
  cos x 2xdx
2
2
2
x
cos
x
dx

cos
x
xdx


2
2
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
2
Evaluate:
We will rearrange the integral to get an exact match:
 
 
1
  cos x 2xdx
2
2
2
x
cos
x
dx

cos
x
xdx


2
We put in a 2
so the pattern
will match.
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
2
Evaluate:
We will rearrange the integral to get an exact match:
 
 
1
  cos x 2xdx
2
2
2
x
cos
x
dx

cos
x
xdx


2
So we must also put in
a 1/2 to keep the
problem the same.
We put in a 2
so the pattern
will match.
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
Evaluate:
We will rearrange the integral to get an exact match:
 
 
1
  cos x 2xdx
2
2
2
x
cos
x
dx

cos
x
xdx


2
2
Using the Chain Rule,
we know that:
   2x cos x 
d
sin x 2
dx
Evaluate:
We will rearrange the integral to get an exact match:
 
 
1
  cos x 2xdx
2
2
2
x
cos
x
dx

cos
x
xdx


2
 
1
  sin x 2   C
2
2
Check Answer:
Check Answer:
Check:
Check Answer:
Check:
From the
chain rule
Indefinite Integrals by Substitution
1) Choose u.
Indefinite Integrals by Substitution
1) Choose u.
2) Calculate du.

du
du 
dx
dx
Indefinite Integrals by Substitution
1) Choose u.
2) Calculate du.
du
du 
dx
dx
3) Substitute u.
Arrange to have du in your integral also.
(All xs and dxs must be replaced!)

Indefinite Integrals by Substitution
1) Choose u.
2) Calculate du.
du
du 
dx
dx
3) Substitute u.
Arrange to have du in your integral also.
(All xs and dxs must be replaced!)

4) Solve the new integral.
Indefinite Integrals by Substitution
1) Choose u.
2) Calculate du.
du
du 
dx
dx
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.
Example
A linear substitution:
Let u = 3x + 2. Then du = 3dx.
e
3x  2
1 3x  2
dx   e
3dx
3
1 u
  e du
3
1 u
 e C
3
1 3x  2
 e
C
3
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 u
For
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.
Practice
Let u =
du =
Practice
Let u = x2 + 1
du =
Practice
Let u = x2 + 1
du = 2x dx
Practice
Let u = x2 + 1
du = 2x dx
Practice
Let u = x2 + 1
du = 2x dx
Make this a
2x dx and we’re
set!
Practice
Let u = x2 + 1
du = 2x dx
Practice
Let u = x2 + 1
du = 2x dx
Practice
Let u = x2 + 1
du = 2x dx
Practice
Let u = x2 + 1
du = 2x dx
 12 ln x 2  1  C
Practice
Let u =
du =
Practice
Let u = sin(x)
du =
Practice
Let u = sin(x)
du = cos(x) dx
Practice
Let u = sin(x)
du = cos(x) dx
Practice
Let u = sin(x)
du = cos(x) dx
Practice
Let u = sin(x)
du = cos(x) dx
Practice
An alternate possibility:
Let u =
du =
Practice
An alternate possibility:
Let u = cos(x)
du = –sin(x) dx
Practice
An alternate possibility:
Let u = cos(x)
du = –sin(x) dx
Practice
An alternate possibility:
Let u = cos(x)
du = –sin(x) dx
Practice
An alternate possibility:
Let u = cos(x)
du = –sin(x) dx
Practice
Note:
Practice
Note:
Practice
Note:
What’s the difference?
Practice
Note:
What’s the difference?
Practice
Note:
What’s the difference?
Practice
Note:
What’s the difference?
This is 1!
Practice
Note:
What’s the difference?
Practice
Note:
What’s the difference?
That is, the difference is a constant.
In-Class Assignment
 3x  1 dx
2

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?