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The Evaluation Theorem Section 5. The Evaluation Theorem: If f is continuous on the interval (^) a , b then
a
b f x dx F b − F a
where F ′^ f (i.e. F is an antiderivative of f ).
Notation:
a
b f x dx F x | a b F b − F a
Examples:
− 2
3 4 x 3 dx 25
0
2 7 x^2 dx (^563)
0
sin x dx 2
− 1
1 e x^ dx e − (^1) e e
e
1
2 1 x dx^ ^ ln 2
1
9 x − 1 x
dx (^403)
PROBLEMS:
7. What’s wrong?
− 1
2 2 x^3
dx
− 1
2 2 x −^3 dx 2 x
− 2 − (^2) − 1
2 − 1 x^2 − 1
2 − 1 4
Integrand is NOT continuous! Show graphically. Cannot evaluate this integral. (It is what we call divergent. We will discuss this more in Section 5.10.)
8. Can we integrate?
0
1 e x
2 dx
No! There is no elementary function that is an antiderivative of the integrand. Must approximate via numerical methods.
9. What about
0
1 x 3 x^2 4 dx
Yes. Via the Substitution Method (Section 5.5).
10. What about
0
2 | x^2 2 x − 3| dx
Yes. Note first that x^2 2 x − 3 x 3 x − 1 Thus
| x^2 2 x − 3|
x^2 2 x − 3 if x ≤ −3 or x ≥ 1
− x^2 − 2 x 3 if − 3 x 1
-4 -3 -2 -1 1 2
2
4
x
y
Graph of x^2 2 x − 3
-4 -3 -2 -1 1 2
2
4
x
y
Graph of | x^2 2 x − 3|
0
2
| x^2 ^2 x^ −^ 3| dx^ ^
0
1
− x^2 2 x − 3 dx
1
2 x^2 2 x − 3 dx
Total Change Theorem ( Application ) The integral of a rate of change of a quantity over some interval is the total change of the quantity
over that interval.
a
b F ′ x dx F b − F a
Examples:
1. If V t is the volume (in gallons) of water in a reservoir at time t (in minutes) then its derivative V ′ t is the rate (in gals/min) that the water flows into () or out of (−) the reservoir. So
t 1
t 2 V ′ t dt V t 2 − V t 1
is the change in the amount of water in the reservoir between time t 1 and t 2. Example: Water is being pumped into a reservoir at a rate of 2 t^3 gallons per minute where t is measured in minutes. Assume the filling process begins at t 0. What is the increase in water in the reservoir between 1 and 3 minutes after the process starts?
V 3 − V 1
1
3 2 t^3 dt 40 gallons
2. If s t is the position function of an object moving along a straight line, then its velocity is v t s ′ t and
t 1
t 2 v t dt s t 2 − s t 1
