Math 220
November 6
I. Evaluate the integral.
1.
2
Z
(3 − 2x − 3x2 )dx
1
2.
3
Z
e2 dx
0
3.
Z
2
√
1 + x−2/3 + 2 x + 3x5/2 dx
1
4.
4
Z
√
(1 + 2t + 3t2 ) tdt
1
5.
1
Z
t2 dx
0
6.
π
Z
sec2 (x)dx
3π/4
7.
2
Z
1
x2 + 3x + 2
2x
8.
1
Z
0
9.
1
Z
0
10.
Z
dx
4
dx
1 + x2
4 + 2x2
dx
1 + x2
π/2
[2 cos(x) + sin(x)]dx
0
11.
Z
π/2
csc2 (x)dx
π/4
12.
Z
π
(2ex + cos(x))dx
0
1
13.
π
Z
(2 sin(x) + sin(2x))dx
0
14.
3
Z
1
15.
x 3
+
3 x
dx
1
Z
(x5 + 5x + e5x + x5e )dx
0
16.
Z
0
1/2
x4 − 1 x3 + 3x2 + 9x + 27
+
x2 − 1
x+3
17.
Z
2
−1
dx
1
dx
x3
II. Find the general indefinite integral.
1.
Z
2.
Z
3.
4.
x2 + x−3 + x−1 +
√
5
x dx
√
√
( x + 1)( 3 x + x)dx
Z
sin(2x)
dx
cos(x)
Z
csc(x)(sin(x) − csc(x))dx
5.
Z
sin(θ) + sin(θ) csc2 (θ)
dθ
cot2 (x)
III. Water flows from the bottom of a storage tank at a rate of r(t) = 600 − 3t2 liters per
minute, where 0 ≤ t ≤ 50. If the tank initially held 3000 liters, how much water would be
left in the tank after 5 minutes.
2
1
Solutions
I. Evaluate the integral.
1.
2
Z
(3 − 2x − 3x2 )dx
1
Answer:
Z
2
(3 − 2x − 3x2 )dx = 3x − x2 − x3
i2
1
1
= (6 − 4 − 8) − (3 − 1 − 1)
= (−6) − 1
= −7
2.
Z
3
e2 dx
0
Answer:
Z
3
e2 dx = e2 x
i3
0
0
2
= 3e − 0
= 3e2
3.
Z
2
√
1 + x−2/3 + 2 x + 3x5/2 dx
1
Answer:
Z
1
2
√
x1/3
x3/2
x7/2 i2
1 + x−2/3 + 2 x + 3x5/2 dx = x +
+2
+3
1/3
3/2
7/2 1
i2
4
6
= x + 3x1/3 + x3/2 + x7/2
3
7
1
√
4√
12 √
4 6
3
= (2 + 3 2 +
8+
128) − (1 + 3 + + )
3
7
3 7
√
4√
12 √
4 6
3
= −2 + 3 2 +
8+
128 − −
3
7
3 7
3
4.
1
Z
t2 dx
0
Answer:
1
Z
t2 dx = t2 x
i1
0
0
2
=t
5.
Z
π
sec2 (x)dx
3π/4
Answer:
Z
π
iπ
sec2 (x)dx = tan(x)
3π/4
3π/4
= tan(π) − tan(3π/4)
= 0 − (−1)
=1
6.
Z
1
2
x2 + 3x + 2
2x
dx
Answer:
Z
1
2
x2 + 3x + 2
2x
2
x 3 1
dx =
+ +
dx
2 2 x
1
i2
x2 3
=
+ x + ln |x|
4
2
1
4 3
1 3
= ( + 2 + ln(2)) − ( + + ln(1))
4 2
4 2
7
= (1 + 3 + ln(2)) − ( )
4
7
= 4 + ln(2) −
4
9
= + ln(2)
4
Z
4
7.
π/2
Z
[2 cos(x) + sin(x)]dx
0
Answer:
Z
π/2
iπ/2
[2 cos(x) + sin(x)]dx = 2 sin(x) − cos(x)
0
0
= (2 sin(π/2) − cos(π/2)) − (2 sin(0) − cos(0))
= (2 − 0) − (0 − 1)
=3
8.
Z
π/2
csc2 (x)dx
π/4
Answer:
Z
π/2
2
csc (x)dx = − cot(x)
iπ/2
π/4
π/4
= (− cot(π/2)) − (− cot(π/4))
= 0 − (−1)
=1
9.
Z
π
(2ex + cos(x))dx
0
Answer:
Z
π
iπ
(2ex + cos(x))dx = 2ex + sin(x)
0
0
π
= 2e + sin(π) − (2e0 + sin(0))
= 2eπ + 0 − (2 + 0)
= 2eπ − 2
10.
Z
π
(2 sin(x) + sin(2x))dx
0
5
Answer:
π
Z
iπ
1
cos(2x)
2
0
1
1
= (−2 cos(π) − cos(2π)) − (−2 cos(0) − cos(0))
2
2
1
1
= (2 − ) − (−2 − )
2
2
=4
(2 sin(x) + sin(2x))dx = −2 cos(x) −
0
11.
Z
3
1
x 3
+
3 x
dx
Answer:
3
Z
1
x 3
+
3 x
dx =
=
=
=
12.
Z
i3
x2
+ 3 ln |x|
6
1
9
1
+ ln(3) − ( + 3 ln(1))
6
6
8
+ ln(3)
6
4
+ ln(3)
3
1
(x5 + 5x + e5x + x5e )dx
0
Answer:
Z
1
x6
5x
e5x
x5e+1 i1
+
+
+
6
ln(5)
5
5e + 1 0
5
1
5
e
1
1
1
=( +
+ +
) − (0 +
+ + 0)
6 ln(5)
5
5e + 1
ln(5) 5
5
−1
4
e
1
=
+
+ +
30
ln(5)
5
5e + 1
(x5 + 5x + e5x + x5e )dx =
0
13.
Z
0
1/2
x4 − 1 x3 + 3x2 + 9x + 27
+
x2 − 1
x+3
6
dx
Answer:
Z
1/2
0
x4 − 1 x3 + 3x2 + 9x + 27
+
x2 − 1
x+3
Z
1/2
(x2 + 1) + (x2 + 3x + 9) dx
dx =
0
Z
=
1/2
(2x2 + 3x + 10) dx
0
i1/2
2x3 3x2
=
+
+ 10x
3
2
0
21 31
=(
+
+ 5) − (0 + 0 + 0)
38 24
1
3
=
+ +5
12 8
II. Find the general indefinite integral.
1.
Z
x2 + x−3 + x−1 +
√
5
x dx
Answer:
Z
2.
x2 + x−3 + x−1 +
Z
√
5
x3 x−2
x6/5
+
+ ln |x| +
+C
3
−2
6/5
x3 x−2
5x6/5
=
−
+ ln |x| +
+C
3
2
6
x dx =
√
√
( x + 1)( 3 x + x)dx
Answer:
Z
√
√
( x + 1)( 3 x + x)dx =
Z
(x5/6 + x1/3 + x3/2 + x)dx
x11/6 x4/3 x5/2 x2
+
+
+
+C
11/6
4/3
5/2
2
6x11/6 3x4/3 2x5/2 x2
=
+
+
+
+C
11
4
5
2
=
7
3.
Z
sin(2x)
dx
cos(x)
Answer:
Z
sin(2x)
dx =
cos(x)
Z
2 sin(x) cos(x)
dx
cos(x)
Z
=
2 sin(x)dx
= −2 cos(x) + C
4.
Z
csc(x)(sin(x) − csc(x))dx
Answer:
Z
Z
csc(x)(sin(x) − csc(x))dx =
(1 − csc2 (x))dx
= x + cot(x) + C
5.
Z
sin(θ) − sin(θ) csc2 (θ)
dθ
cot2 (θ)
Answer:
Z
sin(θ) − sin(θ) csc2 (θ)
dθ =
cot2 (θ)
sin(θ)(1 − csc2 (θ)
dθ
cot2 (θ)
Z
sin(θ)(− cot2 (θ)
=
dθ
cot2 (θ)
Z
= − sin(θ)dθ
Z
= cos(x) + C
III. Water flows from the bottom of a storage tank at a rate of r(t) = 600 − 3t2 liters per
minute, where 0 ≤ t ≤ 50. If the tank initially held 3000 liters, how much water would be
left in the tank after 5 minutes.
Answer:
8
The volume of the water that has left the tank is:
Z
5
r(t)dt
V (t) =
0
Z
5
(600 − 3t2 )dt
0
i5
3
= 600t − t
=
0
= 3000 − 53
= 3000 − 125
= 2875
The volume of the water remaining in the tank is:
3000 - 2875 = 125
125 liters3
9