Post on 04-Jan-2016
The Fundamental Theorem of Calculus
Objective: The use the Fundamental Theorem of Calculus to evaluate
Definite Integrals.
The Fundamental Theorem of Calculus Pt. 1
• If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then
• This is also expressed in this form.
)()()( aFbFdxxfb
a
bab
a
xFdxxf )()(
Example 1
• Evaluate 2
1
xdx
Example 1
• Evaluate
2
3)1(
2
1)2(
2
1
2
1 222
1
22
1
xxdx
2
1
xdx
Example 2
• Evaluate 3
0
2 )9( dxx
Example 2
• Evaluate 3
0
2 )9( dxx
180)3
2727(
39)9(
3
0
33
0
2
x
xdxx
Example 3
a) Find the area under the curve y = cosx over the interval [0, /2]
Example 3
a) Find the area under the curve y = cosx over the interval [0, /2]
• Since the cosx > 0 over the interval [0, /2] the area under the curve is
10sin
2sinsincos 2/
0
2/
0
xxdxA
Example 3
b) Make a conjecture about the value of the integral
0
cos xdx
Example 3
b) Make a conjecture about the value of the integral
• The given integral can be interpreted as the signed area between the graph of y = cosx and the interval [0, ]. We can see that the area above and below the x-axis is the same. The value is zero.
00sinsinsincos 0
0
xxdx
Example 4
• Evaluate 9
1
dxx
Example 4
• Evaluate 9
1
dxx
3
52
3
218)1(
3
2)9(
3
2
3
2 2/32/39
1
2/39
1
xdxx
Example 5
• Here are a few more examples.
Example 5
• Here are a few more examples.
Example 6
• Remember the two theorems we had earlier.
03
1
3
1
3
1
1
31
1
2
x
dxx
82
0
2
16
2
4
0
24
0
0
4
x
xdxxdx
Example 7
• Evaluate if 6
0
)( dxxf2
2
,23
2
x
x
x
x
Example 7
• Evaluate if 6
0
)( dxxf2
2
,23
2
x
x
x
x
6
2
2
0
6
0
)()()( dxxfdxxfdxxf
3
128)242(0
3
82
2
3
3)23(
6
2
2
2
0
36
2
2
0
2
xxx
dxxdxx
Total Area
• If f is a continuous function on the interval [a, b], then we define the total area between the curve y = f(x) and the interval [a, b] to be
|)(|b
a
dxxfTotalArea
Total Area
• To compute total area, begin by dividing the interval of integration into subintervals on which f(x) does not change sign. On the subintervals on which 0 < f(x) replace |f(x)| with f(x), and on the subintervals for which f(x) < 0 replace |f(x)| with –f(x). Adding the resulting integrals will give total area.
Example 8
• Find the total area between the curve y = 1 – x2 and the x-axis over the interval [0, 2].
Example 8
• Find the total area between the curve y = 1 – x2 and the x-axis over the interval [0, 2].
• The area is given by
2
0
2 |1| dxx
2
1
21
0
2 )1()1( dxxdxx
23
4
3
2
33
2
1
31
0
3
xx
xx
Dummy Variables
• The variable used in integration is not important. As long as everything is in terms of one variable, it can be anything you want.
• Because the variable of integration in a definite integral plays no role in the end result, it is often referred to as a dummy variable. You can change the variable of integration without changing the value of the integral.
The Mean-Value Theorem for Integrals
• Let f be a continuous function on [a, b], and let m and M be the minimum and maximum values of f(x) on this interval. Consider the rectangles of heights m and M over the interval [a, b]. It is clear geometrically that the area under f(x) is at least as large as the rectangle of height m and no larger than the area of the rectangle of height M.
The Mean-Value Theorem for Integrals
• Theorem 6.6.2• If f is continuous on a closed interval [a, b], then there
is at least one point x* in [a, b] such that
))(()( * abxfdxxfb
a
Example 9
• Since f(x) = x2 is continuous on the interval [1, 4], the Mean-Value Theorem for Integrals guarantees that there is a point x* in [1, 4] such that
2**4
1
2 )(3)14)(( xxfdxx
213
4
1
34
1
2
x
dxx
21)(3 2* x 7)( 2* x 7* x
Theorem 6.6.3
• The Fundamental Theorem of Calculus, Part 2• If f is continuous on an interval I, then f has an
antiderivative on I. In particular if a is any point in I, then the function f defined by
• is an antiderivative of f on I; that is, for each x in I, or in an alternative notation
)()(/ xfxF
x
a
dttfxF )()(
)()( xfdttfdx
d x
a
Example 10
• Find
x
dttdx
d
1
3
Example 10
• Find
x
dttdx
d
1
3
3
1
3 xdttdx
d x
Example 10
• Find
• Let’s integrate and the take the derivative to confirm this result.
3
1
3 xdttdx
d x
x
dttdx
d
1
3
4
1
44
4
1
4
1
3
xtdtt
xx3
4
4
1
4x
x
dx
d
Chain Rule
• The Fundamental Theorem of Calculus, Part 2 also has a form the applies the chain rule.
23
1
3sin6sin
2
xxtdtdx
d x
duufdttfdx
d u
a
)()(
Chain Rule
• Let’s integrate, then take the derivative to confirm this rule.
23
1
3sin6sin
2
xxtdtdx
d x
duufdttf
dx
d u
a
)()(
1cos3coscossin 231
3
1
2
2
xttdt xx
063sin1cos3cos 22 xxxdx
d
Integrating Rates of Change
• The Fundamental Theorem of Calculus has a useful interpretation that can be seen by rewriting it in a slightly different form. Since F is an antiderivative of f on the interval [a, b], we can use the relationship F/(x) = f(x) to rewrite it like this:
)()()(/ aFbFdxxFb
a
Integrating Rates of Change
• In this formula we can view F/(x) as the rate of change of F(x) with respect to x, and we can view F(b) – F(a) as the change in the value of F(x) as x increases from a to b. This leads us to the following principle
• Integrating Rate of Change- Integrating the rate of change of F(x) with respect to x over an interval [a, b] produces the amount of change in the value of F(x) that occurs as x increases from a to b .
)()()(/ aFbFdxxFb
a
Integrating Rates of Change
• Here are some examples of this idea:1. If P(t) is a population at time t, the P/(t) is the rate at
which the population is changing at time t and
is the amount of change in the population between times t1 and t2.
)()()( 12/
2
1
tPtPdtxPt
t
Integrating Rates of Change
• Here are some examples of this idea:2. If A(t) is the area of an oil spill at time t, and A/(t) is
the rate at which the area of the spill is changing at time t, and
is the amount of change in the area of the spill between times t1 and t2.
)()()( 12/
2
1
tAtAdtxAt
t
Homework
• Section 5.6
• 5-33 odd
• 59