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

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)

Chapter 5.3 & 5.5

February 6, 2007

Fundamental Theorem of Calculus (Part 1)

d

drv eu2 v −1 dv

1

r3

∫    

Fundamental Theorem of Calculus (Part 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

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

Example

x3dx2

3

∫

Example

Evaluate:

ex  dx1

3

∫

1+ 3y−y2( ) dy0

4

∫

t

t +  t23 dt

1

64

∫

Evaluate:

r−5dr−2

3

∫

sec2 θdθ0

π4∫

6

1−y2 dy1

2

32∫

Evaluate:f (x) dx

−π

π

∫ ,         f(x) =x       −π ≤x≤0sinx    0 ≤x≤π⎧⎨⎩

Given:

f (x) =

0         x< 0x         0 ≤x≤12 −x    1 < x≤20          x> 2

⎧

⎨⎪⎪

⎩⎪⎪

Write a similar expression for g(x) = f(t)dt0

x

∫

Evaluate:

Multiply out:

2x +1( )2 dx−1

2

∫

= 4x2 + 4x + 1( )dx−1

2

∫

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

What if instead? 2x +1( )10 dx∫

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

The same substitution holds for the higher power!

With

2x +1( )10 dx∫

u =2x+1du =2dx

1

22x +1( )10 2dx∫

Our Original Exampleof a Definite Integral:

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

2x +1( )2 dx−1

2

∫

u =2x+1

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)

∫

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