HOMEWORK SOLUTIONS:
Section 4.9
13.)
√
u4 + 3 u
f (u) =
u2
Splitting the fraction,
√
u4
3 u
+
u2
u2
1
u2
= u2 + 3 2
u
3
= u2 + 3u− 2 .
f (u) =
So the antiderivative is
1
1 3 3u− 2
u +
+C
3
− 21
1
1
= u3 − 6u− 2 + C
3
F (u) =
14.) By examining each part,
F (u) = 3ex + 7 tan x + C.
23.) Taking the anti-derivative of f 00 , we have
f 0 (x) = 3x2 + 4x3 + C.
Taking the anti-derivative again, we have
f (x) = x3 + x4 + Cx + D.
where C and D are constants.
30.) Here, we first take the antiderivative of f 0 (x):
f (x) = 2x4 + 6x2 + 3x + C.
Now, we use the fact that f (1) = 6 to find C:
f (1) = 2(1) + 6(1) + 3(1) + C = 6
11 + C = 6
C = −5.
1
So
f (x) = 2x4 + 6x2 + 3x − 5.
31.) Here,
1
1
3
f 0 (x) = x 2 (6 + 5x) = 6x 2 + 5x 2 .
So
f (x) =
6
3
2
3
x2 +
5
5
2
3
5
x2 + C
5
= 4x 2 + 2x 2 + C
Now, we use the fact that f (1) = 10 to find C:
f (1) = 4(1) + 2(1) + C = 10
C = 4.
So
3
5
f (x) = 4x 2 + 2x 2 + 4.
44.) First, we take the anti-derivative of f 00 (t) to find f 0 (t):
f 0 (t) = 2et − 3 cos t + C.
Next, we take the anti-derivative again to find f (t):
f (t) = 2et − 3 sin t + Ct + D,
where C and D are both constants.
Now, to find C and D, we use the initial conditions. First, let us use the
condition f (0) = 0:
f (0) = 2e0 − 3 sin 0 + C(0) + D = 0
2+D =0
D = −2.
So
f (t) = 2et − 3 sin t + Ct − 2.
Now, we use the condition that f (π) = 0:
f (π) = 2eπ − 3 sin π + C(π) − 2 = 0
2eπ − 2 + Cπ = 0
2 − 2eπ
.
C=
π
Thus,
f (t) = 2et − 3 sin t +
2
2 − 2eπ
t − 2.
π
Section 5.2
2.) We split the interval 0 to 3 into 6 intervals, each with size 12 . So
1
1
3
5
(f ( ) + f (1) + f ( ) + f (2) + f ( ) + f (3)
2
2
2
2
1 3
3
5
= (− − 1 − + 0 + + 3)
2 4
4
4
1 7
= ( )
2 4
7
=
8
R6 =
5.) I think I meant to assign problem 4. Oops.
Well, whatever. For this one, the left endpoints give area 6 and the right
ones give 4. See the diagram once I upload the diagram.
21.) Let’s split this into three intervals (the problem doesn’t specify, so I will).
The total length of the region is 6, so each rectangle has length 2.
So
L3 = 2(f (−1) + f (1) + f (3)) = 2(−2 + 4 + 10) = 24
and
R3 = 2(f (1) + f (3) + f (5)) = 2(4 + 10 + 16) = 38.
24.) Let’s split this into 5 intervals of length 1 apiece:
L5 = 1(f (0) + f (1) + f (2) + f (3) + f (4)) = 1(1 + 3 + 17 + 55 + 129) = 205
and
R5 = 1(f (1) + f (2) + f (3) + f (4) + f (5)) = 1(3 + 17 + 55 + 129 + 251) = 455
3
Section 5.3
25.) Bringing the denominator up,
Z 2
1
3
dt
t4
Z
=
2
3t−4 dt
1
2
= −t−3 1
2
1
=− 3
t 1
1
=−
−1
8
7
= .
8
√
3
28.) Noting that x x = x 2 ,
Z
1
3
(3 + x 2 )dx
0
2 5
= 3x + x 2
5
2
=3+ −0
5
17
=
5
1
0
32.) In this case,
Z
π
4
sec θ tan θdθ
0
π4
= sec θ
0
π
= sec( ) − sec 0
4
1
1
=
π −
cos( ) cos 0
√ 4
= 2−1
4
40.) Here, we split up the fraction:
Z
2
4
u2
+ 3 )du
3
u
u
Z 2
1
=
(4u−3 + )du
u
1
2
−2
= −2u + ln |u|
(
1
1
2
2
= − 2 + ln |u|
u
1
2
= − + ln |2| − [−2 + ln 1]
4
3
= + ln 2
2
since ln 1 = 0.
5
Section 5.4
14.) Splitting the integral up,
Z
(csc2 t − 2et )dt
Z
Z
= csc2 tdt − 2et dt
= − cot x − 2et + C
6
Section 5.5
13.) In the case of
Z
dx
,
5 − 3x
let us take
u = 5 − 3x.
So
du = −3dx
du
dx =
−3
Plugging back in:
Z
du
−3(u)
Z
1
du
=−
3
u
1
= − ln |u| + C
3
1
= − ln |5 − 3x| + C.
3
33.) Here, we note that cot x appears to be the ”inside” function. So we take
u = cot x.
Then
du = − csc2 xdx
du
dx = − 2
csc x
Plugging back in:
Z √
cot x csc2 x
Z
√
du
=
u csc2 x(
)
− csc2 x
Z
√
=−
udu
2 3
= − u2 + C
3
3
2
= − (cot x) 2 + C.
3
7