6. Integration
6.5: Antiderivatives
More Properties of Definite Integrals
Ra
1. a f (x) dx =
2.
Rb
3.
Ra
(f (x) ± g(x)) dx =
a
4.
Rb
5.
Rb
a
a
a
cf (x) dx =
f (x) dx =
f (x) dx =
Example 1. Find the area represented by
a)
Z
1
x dx.
−2
b) Find the antiderivative F (x) of f (x) = x, and evalute F (1) − F (−2).
1
The Fundamental Theorem
of Calculus, Part I (FTC I): If f is continuous on [a, b], then the
Rx
function g defined by a f (t) dt is continuous on [a, b] and differentiable on (a, b), and
g 0 (x) = f (x).
Example 2. Find the derivative of each function
a)
Z x
(t2 − 1)2 dt;
g(x) =
1
b)
Z
u
g(u) =
t3
1
1
dt;
+1
The Fundamental Theorem of Calculus, Part II (FTC II): Suppose f is a continuous function
defined on a closed interval [a, b], then
Z
b
f (x) dx = F (b) − F (a)
a
where F is any antiderivative of f .
Example 3. Evaluate the following integrals:
a)
Z
3
e3x dx
−2
b)
Z
B√
6
2x + 5 dx,
−1
where B is a real number and B > −1
2
c)
Z
4
e3x−4 dx
2
d) Find the average value of the function f (x) = 3x2 − x on the interval [1, 3]
e)
Z
1
2
1
dx
(−x + 5)3
3
f)
Z
1
0
x2
Z
5
x+3
dx
+ 4x + 3
g)
| − x + 7| dx
−5
4
6.5: Fundamental Theorem of Calculus (continue)
Example 4. Find the area under the parabola y = −(x − 1)2 + 1 for x in the interval [0,2].
NOTE: BecauseR of the relation given by the FTC between the antiderivatives and integrals, the indefinite integral f (x) dx as a function is traditionally used for an antiderivative of f .
The connection between definite integral (as a scalar) with the indefinite integral (as a function)
is given by FTC: If f is continuous on [a, b] then
b
Z
b
Z
f (x) dx
f (x) dx =
a
.
a
Example 5. (Stewart) Suppose an object is moving along a straight line with position function s(t),
velocity function v(t) and acceleration function a(t). Since s0 (t) = v(t), the FTC gives
Z
t2
v(t) dx = s(t2 ) − s(t1 ),
t1
which is the displacement of the object during the time period [t1 , t2 ]. Hence, we can compute the
displacement by integration if the velocity function is known.
Similarly, since v 0 (t) = a(t), the FTC gives
Z t2
a(t) dx = v(t2 ) − v(t1 ),
t1
which is the change in velocity during the time period [t1 , t2 ].
If we want to calculate the distance traveled during the time interval, we have to consider the intervals
when v(t) ≥ 0 (the object moves to the right) and also the intervals when v(t) ≤ 0 (the object moves
to the left). In both cases, the distance is computed by integrating |v(t)|, the speed. Therefore,
Z t2
|v(t)| dx = total distance traveled.
t1
5
Example 6. An object moves along a line so that its velocity at time t is v(t) = 8t + 1 where position
is given in meters and t is given in seconds.
a) Find the displacement of the particle during the time period [0, 2].
b) Find the distance traveled during the this time period.
Finding Approximations of Definite Integrals on the Calculator: (Bollinger) If f is a continous
Rb
function on [a, b], then the value a f (x) dx can be estimated by the following command
M AT H → 9 : f nInt → f nInt(f (x), x, a, b).
Average value of a function
DEFINITION: The average value of a continuous function f on the interval [a, b] is given by
1
b−a
b
Z
f (x) dx.
a
Example 7. Find the average value of the following function on the given interval.
a) f (x) = 9 on [1,10].
b) f (x) = 2x2 + x − 5 on [-1,1].
6