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5.4 The Fundamental Theorem
The Fundamental Theorem of Calculus, Part 1
If f is continuous on , then the function ,a b
x
aF x f t dt
has a derivative at every point in , and ,a b
x
a
dF df t dt f x
dx dx
x
a
df t dt f x
dx
First Fundamental Theorem:
1. Derivative of an integral.
a
xdf t dt
xf x
d
2. Derivative matches upper limit of integration.
First Fundamental Theorem:
1. Derivative of an integral.
a
xdf t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
First Fundamental Theorem:
x
a
df t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
New variable.
First Fundamental Theorem:
cos xd
t dtdx cos x 1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
sinxd
tdx
sin sind
xdx
0
sind
xdx
cos x
The long way:First Fundamental Theorem:
20
1
1+t
xddt
dx 2
1
1 x
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
2
0cos
xdt dt
dx
2 2cosd
x xdx
2cos 2x x
22 cosx x
The upper limit of integration does not match the derivative, but we could use the chain rule.
53 sin
x
dt t dt
dx The lower limit of integration is not a constant, but the upper limit is.
53 sin
xdt t dt
dx
3 sinx x
We can change the sign of the integral and reverse the limits.
The Fundamental Theorem of Calculus, Part 2
If f is continuous at every point of , and if
F is any antiderivative of f on , then
,a b
b
af x dx F b F a
,a b
(Also called the Integral Evaluation Theorem)
We already know this!
To evaluate an integral, take the anti-derivatives and subtract.
Example 10:
24
0tan sec x x dx
The technique is a little different for definite integrals.
Let tanu x2sec du x dx
0 tan 0 0u
tan 14 4
u
1
0 u du
We can find new limits, and then we don’t have to substitute back.
new limit
new limit
12
0
1
2u
1
2We could have substituted back and used the original limits.
Example 8:
24
0tan sec x x dx
Let tanu x
2sec du x dx4
0 u du
Wrong!The limits don’t match!
42
0
1tan
2x
2
21 1tan tan 0
2 4 2
2 21 11 0
2 2
u du21
2u
1
2
Using the original limits:
Leave the limits out until you substitute back.
This is usually more work than finding new limits
Example 9 as a definite integral:
1 2 3
13 x 1 x dx
3Let 1u x 23 du x dx
11 3 22
1( 1) 3xx dx
133 2
1
2( 1)
3x
33 3/ 2 3/ 22(1 1) ( 1 1)
3
22 2
3 4 2
3
Rewrite in form of
∫undu
No constants needed- just integrate using the power rule.
Substitution with definite integrals
7
0
4 3xdx4 3
3
3
u x
du dx
dudx
71 3 3 3 3
2 2 2 2 2
0
1 1 2 2 2 234(4 3 ) (25 4 )
3 3 3 9 9 9u du u C x
2525 1 3 3 3
2 2 2 2
4 4
1 2 2 234(25 4 )
3 9 9 9u du u
Using a change in limits
