Section 5.4Theorems About Definite Integrals

Properties of Limits of Integration

• If a, b, and c are any numbers and f is a continuous function, then

b

a

a

bdxxfdxxf )()(

b

a

b

c

c

adxxfdxxfdxxf )()()(

Properties of Sums and Constant Multiples of the Integrand

• Let f and g be continuous functions and let c be a constant, then

b

a

b

a

b

a

b

a

b

a

dxxfcdxxcf

dxxgdxxfdxxgxf

)()()2

)()())()(()1

Example• Given that

find the following:

cbadxxfdxxfc

a

b

a and,10)(,5)(

c

bdxxf )(

b

cdxxf )(

c

adxxf )(3

Using Symmetry to Evaluate Integrals

• An EVEN function is symmetric about the y-axis

• An ODD function is symmetric about the originIf f is EVEN, then

If f is ODD, then

aa

adxxfdxxf

0)(2)(

0)( a

adxxf

EXAMPLE

Given that

Find

2sin0

xdx

dxxb

xdxa

|sin|__.

sin1__.

Comparison of Definite Integrals• Let f and g be continuous functions

)()()(then

,for)(If1)

abMdxxfabm

bxaMxfmb

a

b

a

b

adxxgdxxf

thenbxaforxgxf

)()(

,)()(If)2

Example

• Explain why

03)cos( dxx

The Area Between Two Curves

• If the graph of f(x) lies above the graph of g(x) on [a,b], then

b

adxxgxf ))()((

Area between f and g on [a,b]

Let’s see why this works!

Find the exact value of the area between the graphs of

y = e x + 1 and y = xfor 0 ≤ x ≤ 2

This is the graph of y = e x + 1 What does the integral from 0 to 2 give us?

Now let’s add in the graph of y = x

Now the integral of x from 0 to 2 will give us the area under x

So if we take the area under e x + 1 and subtract out the area under x, we get the area between the 2 curves

So we find the exact value of the area between the graphs of

y = e x + 1 and y = xfor 0 ≤ x ≤ 2 with the integral

2 2 2

0 0 0( 1) ( 1 )x xe dx xdx e x dx

Notice that it is the function that was on top minus the function that was on bottom

Find the exact value of the area between the graphs of

y = x + 1 and y = 7 - x for 0 ≤ x ≤ 4

Let’s shade in the area we are looking for

Notice that these graphs switch top and bottom at their intersectionThus we must split of the integral at the intersection point and switch the order

So to find the exact value of the area between the graphs of

y = x + 1 and y = 7 - x for 0 ≤ x ≤ 4 we can use the following integral

3 4

0 3

3 4

0 3

(7 ) ( 1) ( 1) (7 )

(6 2 ) (2 6)

x x dx x x dx

x dx x dx

Find the exact value of the area enclosed by the graphs of

y = x2 and y = 2 - x2

Let’s shade in the area we are looking for

In this case we weren’t given limits of integrationSince they enclose an area, we use their intersection points for the limits

So to find the exact value of the area enclosed by the graphs of

y = x2 and y = 2 - x2 we can use the following integral

1 12 2 2

1 1(2 ) ( ) (2 2 )x x dx x dx