Section 5.4 Theorems About Definite Integrals. Properties of Limits of Integration If a, b, and c...
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Transcript of Section 5.4 Theorems About Definite Integrals. Properties of Limits of Integration If a, b, and c...
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