Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Integral Calculus - Exercises
6
...

The Indefinite Integral

In problems 1 through 7, find the indicated integral
...

xdx
Solution
...

3
3
R x
2
...

Z
Z
x
3e dx = 3 ex dx = 3ex + C
...


4
...

Z
Z
Z
√
√ Z √
2
2
(3x − 5x + 2)dx = 3 x dx − 5
xdx + 2 dx =

R³

√ 2 √
1
= 3 · x3 − 5 · x x + 2x + C =
3
3
√
2
= x3 − x 5x + 2x + C
...

¶
Z
Z
Z µ
Z
1
2
1
1
3
1
−2
− 2+√
dx − 2 x dx + 3 x− 2 dx =
dx =
2x x
2
x
x
1
1
=
ln |x| − 2 · (−1)x−1 + 3 · 2x 2 + C =
2
√
ln |x| 2
=
+ + 6 x + C
...


6
...

¶
Z
Z µ
Z
Z
6
1
x
x
2e + + ln 2 dx = 2 e dx + 6
dx + ln 2 dx =
x
x
= 2ex + 6 ln |x| + (ln 2)x + C
...

Z
Z
Z 2
Z
3
1
1
x + 3x − 2
√
dx =
x 2 dx + 3 x 2 dx − 2 x− 2 dx =
x
2 5
2 3
1
=
x 2 + 3 · x 2 − 2 · 2x 2 + C =
5
3
3
1
2 5
=
x 2 + 2x 2 − 4x 2 + C =
5
√
√
2 2√
=
x x + 2x x − 4 x + C
...
(x3 − 2x2 ) x1 − 5 dx
Solution
...

4
3

8
...

Solution
...
Thus
f 0 (x) = x3 −

2
+2
x2

and so f(x) is the indefinite integral
¶
Z µ
Z
2
0
3
x − 2 + 2 dx =
f (x) =
f (x)dx =
x
1 4 2
=
x + + 2x + C
...

4
4
Therefore, the desired function is f (x) = 14 x4 + x2 + 2x − 54
...
It is estimated that t years from now the population of a certain lakeside
community will be changing at the rate of 0
...
2t + 0
...
Environmentalists have found that the level of pollution in the lake increases at the rate of approximately 5 units per 1000
people
...
Let P (t) denote the population of the community t years
from now
...
6t2 + 0
...
5
...
6t2 + 0
...
5
...
6t2 + 0
...
5)dt =
= 0
...
1t2 + 0
...
During the next 2 years, the population will grow
on behalf of
P (2) − P (0) = 0
...
1 · 22 + 0
...
6 + 0
...

Hence, the pollution in the lake will increase on behalf of 5 · 3 = 15
units
...
An object is moving so that its speed after t minutes is v(t) = 1+4t+3t2
meters per minute
...
Let s(t) denote the displacement of the car after t minutes
...

During the 3rd minute, the object travels

s(3) − s(2) = 3 + 2 · 9 + 27 + C − 2 − 2 · 4 − 8 − C =
= 30 meters
...
Check your answers
by differentiation
...
R x5 dx
2
...

dx
4
...

3
x
−
+
5
...

x − 2√x + 2 dx
7
...

− 2x32 + e2 + 2x dx 10
...
R x 2x + x dx
12
...

x(2x + 1) dx
14
...


15
...

16
...

17
...
If the current
population is 10000, what will the population be 8 months from now?
18
...
1t + 0
...
If the current level of
carbon monoxide in the air is 3
...
After its brakes are applied, a certain car decelerates at the constant
rate of 6 meters per second per second
...
)
20
...
If it is traveling at 90 kilometers per hour (25
meters per second) when the brakes are applied, its stopping distance
is 50 meters
...

1
...
− x1 + C
3
5
5
...
12 ex + 25 x 2 + C
3
3
9
...
25 x5 + 13 x3 + C
13
...
f (x) = 2x2 + x − 1
15
...
f (x) = 13 x3 − 52 x2 + 4x; not unique
17
...
4
...
75 meters
20
...
25
(b) 42 meters
(c) 120
...
47 x 4 + C
4
...
2xp2 + x12 + ln |x| + C
√
√
8
...
x − x1 + 2 ln x + C
7
3
12
...
2

45

Integration by Substitution

In problems 1 through 8, find the indicated integral
...
(2x + 6)5 dx
Solution
...

2
12
12
2
...
Substituting u = x − 1 and du = dx, you get
Z
Z
¤
£
5
2
(x − 1) + 3(x − 1) + 5 dx =
(u5 + 3u2 + 5)du =

1 6
u + u3 + 5u + C =
6
1
(x − 1)6 + (x − 1)3 + 5(x − 1) + C
...

6
3
...


5
...
Substituting u = x2 and 12 du = xdx, you get
Z
Z
1
1
1 2
x2
xe dx =
eu du = eu + C = ex + C
...
Substituting u = 1 − x6 and − 16 du = x5 dx, you get
Z
Z
1
1
1
6
5 1−x6
xe
dx = −
eu du = − eu + C = − e1−x + C
...
Substituting u = x5 + 1 and 25 du = 2x4 dx, you get
Z

2x4
2
dx =
5
x +1
5

Z

¯
1
2
2 ¯
du = ln |u| + C = ln ¯x5 + 1¯ + C
...


R

46

3
√10x −5x dx
x4 −x2 +6

Solution
...

7
...
Substituting u = ln x and du = x1 dx, you get
Z
Z
1
1
dx =
du = ln |u| + C = ln |ln x| + C
...


R

ln x2
dx
x

Solution
...

2

9
...

Solution
...

=
11
10
=

10
...

Solution
...

5
25
Homework
In problems 1 through 18, find the indicated integral and check your answer
by differentiation
...
R e5x dx
2
...

dx
4
...

2xe
dx
6
...

3x 2
8
...

10
...
R (3x2 − 1)ex −x dx 12
...

14
...

dx
16
...

dx
18
...
R √
dx
20
...
R x−1
x
x+3
21
...

dx
6 dx
(x−4)2
R (x−5)
x
23
...
Find the function whose tangent has slope x x2 + 5 for each value of
x and whose graph passes through the point (2, 10)
...
Find the function whose tangent has slope 1−3x
2 for each value of x and
whose graph passes through the point (0, 5)
...
A tree has been transplanted and after x years is growing at the rate
1
of 1 + (x+1)
2 meters per year
...
How tall was it when it was transplanted?
27
...
02t million per year
...

1
...
13 ln |3x + 5| + C
2
5
...
(x2 + 8) x2 + 8 + C
9
...
ex −x + C
3
13
...
2 ln 5x + C
17
...
x + ln |x − 1| + C
1
1
21
...

x − 4 ln |2x + 1| + C
4

√
2
...
−e1−x + C
1
6
...
21
(x3 + 1) 4 + C
1
10
...
35 ln |x5 + 5x4 + 10x + 12| + C
14
...
− ln
+C
√x
x
18
...
25 (x + 1)2 x + 1 − 23 (x + 1) x + 1 + C
7
22
...
f (x) = 13 (x2 + 5) x2 + 5 + 1
25
...


7
3

meters

27
...
3

49

Integration by Parts

In problems 1 through 9, use integration by parts to find the given integral
...
xe0
...
Since the factor e0
...
1x
Then,

Z

G(x) =

and

e0
...
1x

f (x) = x
...
1x
0
...
1x dx = 10xe0
...
1x + C =
2
...
1x + C
...
Since the factor e−x is easy to integrate and the factor 3−2x
is simplified by differentiation, try integration by parts with
g(x) = e−x
Then,
G(x) =

Z

and

e−x dx = −e−x

f (x) = 3 − 2x
...


= (2x − 3)e−x + 2e−x + C = (2x − 1)e−x + C
...
In this case, the factor x is easy to integrate, while the
factor ln x2 is simplified by differentiation
...


Then,
G(x) =

Z

1
xdx = x2
2

and

f 0 (x) =

1
2
2x =
2
x
x

INTEGRAL CALCULUS - EXERCISES

50

and so
Z
Z
Z
1 2
1 2
1 22
2
2
2
x ln x dx =
x ln x −
x dx = x ln x − xdx =
2
2 x
2
¢
¡
1 2
1
1
=
x ln x2 − x2 + C = x2 ln x2 − 1 + C
...
x 1 − xdx
√
Solution
...

g(x) = 1 − x
Then,

G(x) =

Z

√
2
3
1 − xdx = − (1 − x) 2
3

and

f 0 (x) = 1

and so
Z
Z
√
3
3
2
2
2
x 1 − xdx = − x(1 − x) +
(1 − x) 2 dx =
3
3
µ
¶
3
5
2
2
2
2
2
− (1 − x)
= − x(1 − x) +
+C =
3
3
5
3
5
2
4
= − x(1 − x) 2 − (1 − x) 2 + C =
3
15
√
√
4
2
= − x(1 − x) 1 − x − (1 − x)2 1 − x + C
...
(x + 1)(x + 2)6 dx
Solution
...


and

f 0 (x) = 1

and so
Z
Z
1
1
6
7
(x + 1)(x + 2) dx =
(x + 1)(x + 2) −
(x + 2)7 dx =
7
7
1
11
=
(x + 1)(x + 2)7 −
(x + 2)8 + C =
7
78
1
=
[8(x + 1) − (x + 2)] (x + 2)7 + C =
56
1
=
(7x + 6)(x + 2)7 + C
...


51

R

x3 e2x dx
Solution
...


and

f (x) = x2
...

2

and so
R

1
e2x dx = e2x
2

f (x) = x3
...

f 0 (x) = 1

and so
Z

1
1
xe dx = xe2x −
2
2
2x

Z

1
1
e2x dx = xe2x − e2x
...

2
4
4
8

INTEGRAL CALCULUS - EXERCISES
7
...
In this case, the factor x13 is easy to integrate, while the
factor ln x is simplified by differentiation
...


Then,
G(x) =

Z

1
1
1
dx = − x−2 = − 2
3
x
2
2x

and

f 0 (x) =

1
x

and so
¶
µ
Z
Z
ln x
ln x 1
ln x 1
1
1
dx = − 2 +
dx = − 2 +
− 2 +C =
x3
2x
2
x3
2x
2
2x
ln x
1
= − 2 − 2 + C
...
x3 ex dx
³
´
2
x2
Solution
...

Then, from Exercise 6
...
3 you get
Z
1 2
2
G(x) = xex dx = ex
2
and so
Z

9
...

2
3 x2

R

x3 (x2 − 1)10 dx
Solution
...

f 0 (x) = 2x
...

2
22
22
Then
Z

1 2 2
x (x − 1) dx =
x (x − 1)11 −
22
1 2 2
x (x − 1)11 −
=
22
1 2 2
=
x (x − 1)11 −
22
3

2

10

Z
1
2x(x2 − 1)11 dx =
22
1 1 2
(x − 1)12 + C =
22 12
1
(x2 − 1)12 + C
...

a
a
R
(b) Use the formula in part (a) to find x3 e5x dx
...
(a) Since the factor eax is easy to integrate and the
factor xn is simplified by differentiation, try integration by parts
with
g(x) = eax
and
f (x) = xn
...


(b) Apply the formula in part (a) with a = 5 and n = 3 to get
Z
Z
1 3 5x 3
3 5x
x e dx = x e −
x2 e5x dx
...

5
5

INTEGRAL CALCULUS - EXERCISES

54

Once again, apply the formula in part (a) with a = 5 and n = 1
to get
Z
Z
1 5x 1
1
1
5x
xe dx = xe −
e5x dx = xe5x − e5x
5
5
5
25
and so
·
µ
¶¸
Z
1 3 5x 3 1 2 5x 2 1 5x
1 5x
3 5x
+C =
x e dx =
xe −
xe −
xe − e
5
5 5
5 5
25
µ
¶
1
3 2
6
6
3
=
x − x + x−
e5x + C
...

R −x 1 through 16,Ruse integration
x
2
1
...
R xe dx
x
3
...
R (1√− x)ex dx
5
...
R x x − 6dx
√ x dx
7
...

R x
R x+2
√
9
...

x2 e−x dx
R
R 2x+1
11
...
R x3 ex dx
13
...
R x(ln x)2 dx
ln x
15
...

x7 (x4 + 5)8 dx
x2
17
...

√
18
...

t

19
...
Express the distance the object travels as a function of time
...
It is projected that t years from now
√ the population of a certain city
will be changing at the rate of t ln t + 1 thousand people per year
...

1
...
−5(x + 5)e− 5 x + C
5
...
19 x(x + 1)9 − 90
(x + 1)10 + C
1
3
9
...
13 x2 − 23 x + 29 e3x + C
13
...
− x1 (ln x + 1) + C
17
...
f (x) = 14 x2 ln x − 12 − 52 − ln 2

19
...
s(t) = −2(t + 2)e− 2 + 4
20
...
(2x − 4)e 2 x + C
4
...
23 x (x − 6) 2 − 15
(x − 6) 2 + C
1
3
8
...
−(x2 + 2x + 2)e−x + C
2
12
...
2 x ln x − ln x + 12 + C
1 4 4
1
16
...
4

56

The use of Integral tables

In Problems 1 through 5, use one of the integration formulas from a table of
integrals (see Appendix) to find the given integral
...
√x2 +2x−3
Solution
...


and then substitute u = x + 1 and du = dx to get
Z
Z
Z
dx
du
dx
√
p
√
=
=
=
x2 + 2x − 3
u2 − 4
(x + 1)2 − 4
¯
¯
¯
¯
p
√
¯
¯
¯
¯
= ln ¯u + u2 − 4¯ + C = ln ¯x + 1 + (x + 1)2 − 4¯ + C =
¯
¯
√
¯
¯
2
= ln ¯x + 1 + x + 2x − 3¯ + C
...
First, rewrite the integrand as
1
1
1
1
1
=
= 4
=
2
2
2
1 − 6x − 3x
1 − 3(2x + x )
4 − 3(x + 1)
3 3 − (x + 1)2

3
...

8 ¯ 1 − 3x ¯
R

3

(x2 + 1) 2 dx
Solution
...


Aplly appropriate formulas (see Appendix, formulas 9 and 13), to get
Z
¯
√
√
√
x
1 ¯¯
¯
2
2
2
2
2
x x + 1dx = (1 + 2x ) x + 1 − ln ¯x + x + 1¯ + C
8
8

INTEGRAL CALCULUS - EXERCISES

57

and

4
...

2
2
Combine these results, to conclude that
Z
¯
p
√
3
1
3 ¯¯
¯
2
2
2
2
2
(x + 1) dx = x(2x + 5) (x + 1) + ln ¯x + x + 1¯ + C
...
Aplly appropriate formula (see Appendix, formula 23), to get
¯¸
¯
·
Z
Z
¯
dx
1
dx
1
3x 3 ¯¯ 2
−x ¯
= −
= − − − ln ¯− + e ¯ + C =
2
−x
−x
2 − 3e
3
3
2
2
3
−3 + e
¯
¯
¯ 1
x 1 ¯¯ 2 − 3e−x ¯¯
x 1 ¯¯
−x ¯
=
+ ln ¯
+
ln
2
−
3e
+ ln |−3| + C
...


1
2

ln |−3| is a constant, you can write
¯
x 1 ¯¯
dx
−x ¯
=
+
ln
2
−
3e
+ C
...
Aplly the reduction formula (see Appendix, formula 29)
Z
Z
n
n
(ln x) dx = x(ln x) − n (ln x)n−1 dx

to get
Z

3

3

(ln x) dx = x(ln x) − 3

Z

(ln x)2 dx =

µ
¶
Z
2
= x(ln x) − 3 x(ln x) − 2 (ln x)dx =
µ
¶
Z
3
2
= x(ln x) − 3x(ln x) + 6 x ln x − dx =
3

= x(ln x)3 − 3x(ln x)2 + 6x ln x − 6x + C
...

R to dx
3dx
2
...

4x(x−5)
R x(2x−3)
R
√ dx
√ dx
3
...

2 −4
x2 +25
R dx
R 9x
5
...
R 3xdx
2 −9
4dx
7
...

Rx2 −x
9
...

x3 e−x dx

INTEGRAL CALCULUS - EXERCISES

58

Locate a table of integrals and use it to find the integrals in Problems 11
through
R 16
...

12
...

(ln 2x) dx 14
...

16
...
One table of integrals lists the formula
¯
¯
Z
¯ x + px2 ± p2 ¯
dx
¯
¯
p
= ln ¯
¯
¯
¯
p
x2 ± p2

while another table lists
Z
¯
¯
p
dx
¯
¯
p
= ln ¯x + x2 ± p2 ¯
...
The following two formulas appear in a table of integrals:
¯
¯
Z
1 ¯¯ p + x ¯¯
dx
=
ln
p2 − x2
2p ¯ p − x ¯
and
√
¯
¯
Z
¯ a + x −ab ¯
dx
2
¯ (for − ab ≥ 0)
...

R dx
(b) Apply both formulas to the integral 9−4x
2
...

¯ x ¯
¯+C
− 13 ln ¯ 2x−3
¯
¯
√
3
...
14 ln 2+x
+C
2−x
¯ x ¯
1
¯
¯
7
...
3 x − 3 x + 9 e + C
11
...


x (ln 2x)2 − 2x ln 2x + 2x + C
p
15
...
ln ab = ln a − ln b
¯
¯
¯+C
18
...


2
...

6
...

10
...

14
...


¯ x ¯
3
¯ ¯+C
ln
− 20
¯
¯ x−5
q
¯
1
4¯
2
ln ¯x + x − 9 ¯ + C
3
¯√ ¯
¯
¯
1
− 2√3 ln ¯ √3+x
¯+C
3−x
¯ x ¯
−4 ln ¯ x−1 ¯ + C
−x
−(x3 + 3x2 + 6x + 6)ep
+C
√
1
(2x +¯ 4) 2x +¯ 4 + (2x + 4) + C
6
√
¯
¯√
1
√ ln ¯ √2x+5−√5 ¯ + C
3 5 ¯
2x+5+ q
5
¯
¯
2 − 2x + 2 ¯ + C
√1 ln ¯x − 1 +
x
3¯
3

INTEGRAL CALCULUS - EXERCISES

6
...
ln 1 (et − e−t ) dt
2
Solution
...

2

1
−2 =
2

R0

3
...
Substitute u = 2x + 6
...
Hence,
¯6
Z
Z 0
1 6 4
1 5 ¯¯
65
3888
4
(2x + 6) dx =
u du =
u¯ =
−0 =

...


Solution
...
Then 13 du = x2 dx, u(1) = 2, and
u(2) = 9
...

3
2
3 2
3u 2
27 6
54
1 (x + 1)

2
...
Substitute u = ln x
...
Hence,
Z

e

5
...

u

R e2

t ln 2tdt
Solution
...

16
16
16

6
...


R1

x2 e2x dx
R
R
Solution
...

2 2 4
4
4
4
4
0

R5

5−t

te− 20 dt
Solution
...


INTEGRAL CALCULUS - EXERCISES
(a) Show that

Rb

Rc
b

f (x)dx =

Rc

f (x)dx
...

R4
(c) Evaluate 0 (1 + |x − 3|)2 dx
...
(a) By the Newton-Leibniz formula, you have
Z b
Z c
f(x)dx +
f(x)dx = F (b) − F (a) + F (c) − F (b) =
a
b
Z c
= F (c) − F (a) =
f (x)dx
...

2 0 2
2

|x| dx +

Z

1

0

|x| dx =

1 1
+ = 1
...

3
3
3 3
3
9
...
[For example,
f (x) = x2 is even
...


−2

(d) A function f is said to be odd if f(−x) = −f (x)
...

−a

R 12
(e) Evaluate −12 x3 dx
...
(a) Substitute u = −x
...
Hence,
Z b
Z −b
f(−x)dx = −
f (u)du = −F (u)|−b
−a = −F (−b) + F (−a)
...


0

By the part (a), you have
Z 0
f(−x)dx = −F (0) + F (−(−a)) = F (a) − F (0) =
−a
Z a
=
f (x)dx
...


0

(c) Since f(x) = |x| is an even function, you have
Z 1
Z 1
Z 1
¯1
|x| dx = 2
|x| dx = 2
xdx = x2 ¯0 = 1 − 0 = 1
...

3 0
3
2

INTEGRAL CALCULUS - EXERCISES

63

(d) Since f(−x) = −f (x), you can write
Z a
Z 0
Z
f (x)dx = −
f (−x)dx +
−a

−a

a

f(x)dx =

0

= F (0) − F (a) + F (a) − F (0) = 0
...

−12

10
...
3t2 + 0
...
By how much will the
value of the crop increase during the next 5 days if the market price
remains fixed at 3 euros per bushel?
Solution
...
Then
the rate of change of the crop with respect to time is
dQ
= 0
...
6t + 1,
dt
and the amount by which the crop will increase during the next 5 days
is the definite integral
Z 5
¢
¡
¢¯5
¡
Q(5) − Q(0) =
0
...
6t + 1 dx = 0
...
3t2 + t ¯0 =
0

= 12
...
5 + 5 = 25 bushels
...

Homework
In problems
17,
R 1 4 1 through
3
(x − 3x + 1) dx
1
...

(2 + 2t + 3t2 ) dt
2
¢
R3¡
5
...

R04 1
√
9
...

dx
2
R02 x +1
13
...

xe−x dx
R−2
10
17
...
1t dt
0

evaluate
R 0 the 5given 2definite integral
2
...

t − √1t dt
1
R −1 t+1
6
...

(2x − 4)5 dx
1
√
R1 3
10
...

dx
R2e2 x−1
14
...

dx
1
x

INTEGRAL CALCULUS - EXERCISES

64

18
...
By how
much will the population of the town increase over the next 8 months?
19
...
If the demand for oil is currently 30
billion barrels per year, how much oil will be consumed during the next
10 years?
20
...
How far does the object travel during the 2nd
minute?
Results
...
20
5
...
43
23
13
...
152
...
− 72
6
...
76
14
...
98 people

3
...
252
11
...
−3e−2 − e2
19
...
48 billion barrels

4
...
− 16
3
12
...
83
20
...
6

65

Area and Integration

In problems 1 through 9 find the area of the region R
...
R is the triangle with vertices (−4, 0), (2, 0) and (2, 6)
...
From the corresponding graph (Figure 6
...

y
6
4

y=x+4

2

-4

0

-2

2

x

Figure 6
...

Hence,
A=

Z

2

−4

(x + 4)dx =

µ

¶¯2
¯
1 2
x + 4x ¯¯ = (2 + 8) − (8 − 16) = 18
...
R is the region bounded by the curve y = ex , the lines x = 0 and
x = ln 12 , and the x axis
...
Since ln 12 = ln 1 − ln 2 = − ln 2 ' −0
...
2) you see that the region in question is
bellow the line y = ex above the x axis, and extends from x = ln 12 to
x = 0
...
2
...

2
2

3
...

Solution
...
3
...
e
...
Also note that the
line y = −x + 10 intersects the x axis at the point (10, 0)
...
3
...
This suggests that you break R into two subregions, R1
and R2 , as shown in Figure 6
...
In particular,
µ
¶¯2
Z 2
¯
1 3
8
32
2
(x + 4)dx =
A1 =
x + 4x ¯¯ = + 8 =
3
3
3
0
0

and

A2 =

Z

10

(−x + 10)dx =

2

Therefore,

µ

¶¯10
¯
1 2
− x + 10x ¯¯ = −50 + 100 + 2 − 20 = 32
...

3
3

INTEGRAL CALCULUS - EXERCISES

67

4
...

Solution
...
4
...
4
...
Hence,
µ
¶¯3
Z 3
Z 3
¯
2 3
2
2
2
A=
[(x +5)−(−x )]dx =
(2x +5)dx =
x + 5x ¯¯ = 18+15 = 33
...
R is the region bounded by the curves y = x2 − 2x and y = −x2 + 4
...
First make a sketch of the region as shown in Figure 6
...
e
...


to get
The corresponding points (−1, 3) and (2, 0) are the points of intersection
...
5
...
Hence,
Z 2
Z 2
2
2
A =
[(−x + 4) − (x − 2x)]dx =
(−2x2 + 2x + 4)dx =
−1
−1
µ
¶¯2
¯
2
16
2
=
− x3 + x2 + 4x ¯¯ = − + 4 + 8 − − 1 + 4 = 9
...
R is the region bounded by the curves y = x2 and y = x
...
Sketch the region as shown in Figure 6
...
Find the points of
intersection by solving the equations of the two curves simultaneously
to get
√
√
√
3
x2 = x
x2 − x = 0
x(x 2 − 1) = 0
and

x=0

x = 1
...

y
3

y=x2

2

y= x

1

-2

-1

0

1

2

3

x

Figure 6
...

√
Notice that for 0 ≤ x ≤ 1, the graph of y = x lies above that of
y = x2
...

A=
( x − x )dx =
3
3
3 3
3
0
0

(a) R is the region to the right of the y axis that is bounded above by
the curve y = 4 − x2 and below the line y = 3
...


INTEGRAL CALCULUS - EXERCISES

69

Solution
...
e
...

(a) Sketch the region as shown in Figure 6
...

y

y=4-x2

4

y=3
3
2
1

-3

-2

-1

0

1

2

x

Figure 6
...

Notice that for 0 ≤ x ≤ 1, the graph of y = 4 − x2 lies above that
of y = 3
...

3
3
3
0
0
0
(b) Sketch the region as shown in Figure 6
...

y

y=4-x2

4

y=3
3
2
1

-3

-2

-1

0

1

2

x

Figure 6
...

Observe that to the left of x = 1, R is bounded above by the
curve y = 3, while to the right of x = 1, it is bounded by the line
y = 4 − x2
...
8, and apply the integral formula

INTEGRAL CALCULUS - EXERCISES

70

for area to each subregion separately
...

3
3

A = A1 + A2 = 3 +

7
...

Solution
...
9
and find the points of intersection of the curve and the lines by solving
the equations
1
=x
x2

1
x
i
...
x3 = 1
=
2
x
8

and

and

x3 = 8

to get
x=1

and

y

y= x12

x = 2
...
9
...
9
...

A2 =
2
x
8
x 16
2 4
16
16
1
1
Thus, the area of the region R is the sum
A = A1 + A2 =

12
3
=
...
R is the region bounded by the curves y = x3 − 2x2 + 5 and y =
x2 + 4x − 7
...
First make a rough sketch of the two curves as shown in
Figure 6
...
You find the points of intersection by solving the equations
of the two curves simultaneously
x3 − 2x2 + 5 = x2 + 4x − 7
x3 − 3x2 − 4x + 12 = 0
x2 (x − 3) − 4(x − 3) = 0
(x − 3)(x − 2)(x + 2) = 0
to get
x = −2,

and

x=2
y
20

x = 3
...
10
...
However, since the curve
y = x3 − 2x2 + 5 is above y = x2 + 4x − 7 between x = −2 and x = 2,
and since y = x2 + 4x − 7 is above y = x3 − 2x2 + 5 between x = 2
and x = 3, it follows that the area of the region between x = −2 and
x = 2, is

INTEGRAL CALCULUS - EXERCISES

A1 =
=

Z

Z

2
−2
2
−2

µ

£¡
¡

72

¢ ¡
¢¤
x3 − 2x2 + 5 − x2 + 4x − 7 dx =

¢
x3 − 3x2 − 4x + 12 dx =

¶¯2
¯
1 4
3
2
=
x − x − 2x + 12x ¯¯ =
4
−2
= 4 − 8 − 8 + 24 − 4 − 8 + 8 + 24 = 32
and the area of the region between x = 2 and x = 3, is
Z 3
¢ ¡
¢¤
£¡ 2
A2 =
x + 4x − 7 − x3 − 2x2 + 5 dx =
2
Z 3
¢
¡ 3
=
−x + 3x2 + 4x − 12 dx =
2
µ
¶¯3
¯
1 4
3
2
=
− x + x + 2x − 12x ¯¯ =
4
2
81
= − + 27 + 18 − 36 + 4 − 8 − 8 + 24 =
4
81
3
= − + 21 =
...

4
4

Homework
In problems 1 through 20 find the area of the region R
...
R is the triangle bounded by the line y = 4 − 3x and the coordinate
axes
...
R is the rectangle with vertices (1, 0), (−2, 0), (−2, 5) and (1, 5)
...
R is the trapezoid bounded by the lines y = x + 6 and x = 2 and the
coordinate axes
...
R is the region bounded by the curve y = x, the line x = 4, and the
x axis
...
R is the region bounded by the curve y = 4x3 , the line x = 2, and the
x axis
...
R is the region bounded by the curve y = 1 − x2 and the x axis
...
R is the region bounded by the curve y = −x2 − 6x − 5 and the x axis
...
R is the region in the first quadrant bounded by the curve y = 4 − x2
and the lines y = 3x and y = 0
...
R is the region bounded by the curve y = x and the lines y = 2 − x
and y = 0
...
R is the region in the first quadrant that lies under the curve y = 16
x
and that is bounded by this curve and the lines y = x, y = 0, and
x = 8
...
R is the region bounded by the curve y = x2 −2x and the x axis
...
)
12
...

13
...

14
...

15
...

16
...

17
...

18
...

19
...

20
...

Results
...
83
6
...
43
16
...
15
7
...
12
17
...
14
8
...
e − 2
18
...

9
...

19
...
16
10
...
32
3
20

---