Section 3 The Definite Integral and Fundamental Theorem of Calculus
In Problems 1 through 30, evaluate the given definite integral using the fundamental theorem of
calculus.
Solution
Z
1
5.
−1
3t4 dx
Z
¯
3 5 ¯¯1
3t dx =
t
5 ¯−1
−1
3
3
= [ (1)5 ] − [ (−1)5 ]
5
5
6
=
5
Z
7.
1
−1
1
−1
9.
1
0
4
(2u1/3 − u2/3 )du
Z
Z
1
¶¯
µ
(2u
1/3
−u
2/3
¯1
3
3
)du =
2 · u4/3 − u5/3 ¯¯
4
5
−1
3 4/3 3 5/3
3
3
= [ (1) − (1) ] − [ (−1)4/3 − (−1)5/3 ]
2
5
2
5
6
= −
5
e−x (4 − ex )dx
Z
1
0
Z
−x
e
x
(4 − e )dx =
0
1
(4e−x − 1)dx
−x
= (−4e
¯1
¯
− x)¯¯
0
−1
= (−4e
4
= 3−
e
Z
11.
1
0
(x4 + 3x3 + 1)dx
Z
0
Z
13.
2
5
− 1) − (−4)
1
¯
¯1
1 5 3 4
(x + 3x + 1)dx =
x + x + x¯¯
5
4
0
1 3
=
+ +1
5 4
= 1.95
4
3
(2 + 2t + 3t2 )dt
Z
2
5
2
(2 + 2t + 3t )dt
¯5
¯
= 2t + t + t ¯
2
3¯
2
= 144
1
Z
15.
3
1
(1 +
1
1
+ 2 )dx
x x
Z
3
1
1
1
(1 + + 2 )dx =
x x
µ
8
+ ln 3 ≈ 3.7653
3
=
Z
17.
−1
−3
t+1
dt
t3
Z
−1
−3
Z
21.
4
√
0
4
√
0
1
dt =
6t + 1
=
0
1
−1
1
1
+ 3 )dt
2
t
−3 t
µ
¶¯
1
1 ¯¯−1
=
− − 2 ¯
t
2t
−3
1
1
1
= (1 − ) − ( − )
2
3 18
2
=
9
=
Z
Z
t+1
dt =
t3
Z
1 4
(6t + 1)1/2 d(6t + 1)
6 0
¯4
¯
1
1/2 ¯
· 2(6t + 1) ¯
6
0
4
3
p
(x3 + x) x4 + 2x2 + 1dx
Z
0
1
Z
p
(x3 + x) x4 + 2x2 + 1dx =
=
=
=
25.
2
e+1
1
0
q
x(x2 + 1) (x2 + 1)2 dx
Z
=
Z
(
1
dt
6t + 1
Z
23.
¶¯
1 ¯¯3
x + ln x −
x ¯1
1 1 2
(x + 1)2 d(x2 + 1)
2 0
¯1
¯
1 2
3¯
(x + 1) ¯
6
0
1 3
(2 − 13 )
6
7
6
x
dx
x−1
Z
2
e+1
x
dx =
x−1
Z
2
e+1
(1 +
1
)dx
x−1
¯e+1
¯
= [x + ln(x − 1)]¯¯
2
= e
2
Z
27.
e2
1
(ln x)2
dx
x
Z
e2
1
Z
(ln x)2
dx =
x
1
=
29.
1/2
1/3
(ln x)2 d ln x
¯
2
¯e
1
(ln x)3 ¯¯
3
1
8
3
=
Z
e2
e1/x
dx
x2
Z
1/2
1/3
e1/x
dx = −
x2
Z
1/2
1/3
e1/x d
¯1/2
¯
¯
1
x
1/x ¯
= −e
1/3
2
3
= e −e
In Problems 31, 33 and 35, f (x) and g(x) are functions that are continuous on the interval −3 ≤
x ≤ 2 and satisfy
Z
2
−3
Z
f (x)dx = 5
2
−3
Z
g(x)dx = −2
Z
1
f (x)dx = 0
−3
1
−3
g(x)dx = 4
In each case, use this information along with rules for definite integrals to evaluate the indicated
integral.
Z
31.
2
−3
[−2f (x) + 5g(x)]dx
Solution(e.g.5.3.6)
By combining the difference rule and constant multiple rule and substituting the given information,
we find that
Z
2
−3
Z
[−2f (x) + 5g(x)]dx = −2
2
−3
Z
f (x)dx + 5
= −2 · 5 + 5 · (−2)
= −20
Z
33.
4
4
g(x)dx
Solution(e.g.5.3.6)
Z
4
Z
35.
1
2
4
g(x)dx = 0
[3f (x) + 2g(x)]dx
Solution(e.g.5.3.6)
3
2
−3
g(x)dx
Z
1
2
Z
[3f (x) + 2g(x)]dx = 3
2
1
Z
= 3[
Z
f (x)dx + 2
Z
2
−3
f (x)dx −
2
1
1
−3
g(x)dx
Z
f (x)dx] + 2[
2
−3
Z
g(x)dx −
1
−3
g(x)dx]
= 3 · (5 − 0) + 2 · (−2 − 4)
= 3
In Problems 37 and 39, find the area of the region R that lies under the given curve y = f (x) over
the indicated interval a ≤ x ≤ b
37. Under y = x4 , over −1 ≤ x ≤ 2
Solution(e.g.5.3.2)
Since f (x) =Zx4 satisfies f (x) ≥ 0 on the interval −1 ≤ x ≤ 2, the area is given by the definite
integral A =
2
−1
x4 dx. That is
Z
¯
1 ¯2
x dx = x5 ¯¯
5 −1
−1
1 5
33
(2 + 1) =
5
5
A =
=
2
4
39. Under y = e2x , over 0 ≤ x ≤ ln 3
Solution(e.g.5.3.2)
Since f (x) =Ze2x satisfies f (x) ≥ 0 on the interval 0 ≤ x ≤ ln 3, the area is given by the definite
ln 3
e2x dx. That is
integral A =
0
Z
A =
=
=
ln 3
0
e2x dx =
¯
1
2
Z
ln 3
0
e2x d(2x)
1 2x ¯¯ln 3
e
2 ¯0
1 2 ln 3
(e
− 1) = 4
2
49. NET GROWTH OF POPULATION A study indicates that t months from now the
population of a certain town will be growing at the rate of P 0 (t) = 5 + 3t2/3 people per month. By
how much will the population of the town increase over the next 8 months?
Solution(e.g.5.3.9, 5.3.10)
The increase in population is given by the definite integral
Z
P (8) − P (0) =
0
8
Z
P 0 (t)dt =
¯
0
8
(5 + 3t2/3 )dt
¯8
9
= (5t + t5/3 )¯¯
5
0
9
= 40 + · 32 = 97.6 ≈ 98 people
5
4
55. CONCENTRATION OF DRUG The concentration of a drag in a patient’s bloodstream t
−0.33t
hours after an injection is decreasing at the rate C 0 (t) = √
mg/cm3 per hour. By how
0.02t2 + 10
much does the concentration change over the first 4 hours after the injection?
Solution(e.g.5.3.9, 5.3.10)
The net change is given by the definite integral
Z
C(4) − C(0) =
0
4
Z
0
P (t)dt =
Z
0
4
4
√
−0.33t
dt
0.02t2 + 10
1
−0.33
√
d(0.02t2 + 10)
2
0.04 0
0.02t + 10
¯4
¯
0.33
2
1/2 ¯
= −
· 2(0.02t + 10) ¯
0.04
0
0.33
[(0.02(4)2 + 10)1/2 − (0.02(0)2 + 10)1/2 ]
= −
0.02
≈ −0.8283 mg/cm3
=
5
Section 4. Applying Definite Integration: Area Between Curves and Average Value
In Problems 7, 10 and 11, sketch the given region R and then find its area.
7. R is the region bounded by the x axis and the curve y = −x2 + 4x − 3.
Solution(e.g.5.4.1, 5.4.2)
To find where the x axis and the curve intersect, solve the equation simultaneously as follows:
−x2 + 4x − 3 = 0
(x − 1)(x − 3) = 0
x = 1, 3
The region R is bounded above by the curve y = −x2 + 4x − 3 and below by the x axis. The area
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
0
0.5
1
1.5
2
2.5
3
3.5
4
of this region is given by the integral
Z
A =
1
3
(−x2 + 4x − 3)dx
µ
¶¯3
¯
¯
¯
1
=
− x3 + 2x2 − 3x
3
1
= (−9 + 18 − 9) − (−
3
4
=
3
10. R is the region bounded by the curve y =
Solution(e.g.5.4.1, 5.4.2)
1
+ 2 − 3)
1
x
and the lines y = x and y = .
2
x
8
To find where the x axis and the curves intersect, solve the equation simultaneously as follows:
1
= x
x2
x = 1
1
The curve y =
1
and the lines y = x intersect at x = 1.
x2
1
x
=
2
x
8
x = 2
The curve y =
1
x
and the lines y = intersect at x = 2.
2
x
8
x
8
x = 0
x =
The curve y = x and the lines y =
x
intersect at x = 0. The required region R is bounded above
8
1
0.8
0.6
0.4
0.2
0
−0.2
0
0.5
1
1.5
2
2.5
1
x
by y = x and below by y = over the interval 0 ≤ x ≤ 1, and is bounded above by y = 2 and
8
x
x
below by y = over the interval 1 ≤ x ≤ 2. Over the interval 0 ≤ x ≤ 1, the area is
8
Z
1
x
)dx
8
0
¯Z 1
7
7 2 ¯¯
x ¯
=
16
16
0
A1 =
=
(x −
Over the interval 1 ≤ x ≤ 2, the area is
Z
3
1
8
− )dx
2
x
1 x
¶¯
µ
5
1
x ¯¯2
=
=
− −
¯
x 16 1 16
A2 =
(
Therefore the total area is given by the sum
A = A1 + A2 =
12
= 0.75
16
11. R is the region bounded by the curve y = x2 − 2x and y = −x2 + 4.
Solution(e.g.5.4.1, 5.4.2)
2
To find where the curves intersect, solve the equation simultaneously as follows:
x2 − 2x = −x2 + 4
(x + 1)(x − 2) = 0
x = −1, 2
Over the interval −1 ≤ x ≤ 2, the curve y = −x2 + 4 is above the curve y = x2 − 2x, and the
6
y
5
4
3
2
1
0
−1
0
2
x
−1
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
2.5
region enclosed by the curves has area
Z
A =
=
2
−1
Z 2
−1
[(−x2 + 4) − (x2 − 2x)]dx
(−2x2 + 2x + 4)dx
µ
¶¯
¯2
1
1
= −2 x2 − x − 2x ¯¯
3
2
−1
= 9
In Problems 19 and 21, find the average value of the given function f (x) over the specified interval
a ≤ x ≤ b.
19. f (x) = 1 − x2 over −3 ≤ x ≤ 3
Solution(e.g.5.4.5, 5.4.6)
The average value of the function over the interval −3 ≤ x ≤ 3 is given by the integral
Z
V
3
1
(1 − x2 )dx
3 − (−3) −3
¯
1
1 3 ¯¯3
=
(x − x )¯
6
3
−3
= −2
=
21. f (x) = e−x (4 − e2x ) over −1 ≤ x ≤ 1
Solution(e.g.5.4.5, 5.4.6)
3
The average value of the function over the interval −1 ≤ x ≤ 1 is given by the integral
Z
V
1
1
e−x (4 − e2x )dx
1 − (−1) −1
Z
1 1
(4e−x − ex )dx
2 −1
¯1
¯
1
−x
x ¯
(−4e − e )¯
2
−1
2
1
(e − )
3
e
=
=
=
=
28. LORENTZ CURVES Find the Gini index of the Lorentz curve L(x) = x2 .
Solution(e.g.5.4.4)
The Gini index is
Z
G = 2
µ
1
0
(x − x2 )dx
¶¯
x2 x3 ¯¯1
= 2
−
2
3 ¯0
¶
1 1
1
= 2( −
=
2 3
3
35. FOOD PRICES Records indicate that t months after beginning of the year, the price of
ground beef in local supermarkets was
P (t) = 0.09t2 − 0.2t + 1.6
dollars per pound. What was the average price of ground beef during the first 3 months of the
year?
Solution(e.g.5.4.5, 5.4.6)
The average price of ground beef during the first 3 months of the year is given by the integral
Z
V
1 3
=
(0.09t2 − 0.2t + 1.6)dt
3 0
¯3
¯
1
3
2
=
(0.03t − 0.1t + 1.6t)¯¯
3
0
= 1.57$
37. BACTERIAL GROWTH The number of bacteria present in a certain culture after t minutes
of an experiment was Q(t) = 2, 000e0.05t . What was the average number of bacteria present during
the first 5 minutes of the experiment? Solution(e.g.5.4.5, 5.4.6)
The average number of bacteria present during the first 5 minutes of the experiment is given by
the integral
Z
V
1 5
(2000e0.05t )dt
5 0
µ
¶¯
1 2000 0.05t ¯¯5
=
e
¯
5 0.05
0
≈ 2272.2
=
4