Math 132.
1.
A Few Practice Questions from 7.4 — 7.7 and 8.1 — 8.4
To find the arc length of the curve y =
1 5x
e + e−5x for 0 ≤ x ≤ 3, we compute
10
dy
e5x − e5x
=
dx
2
then
e10x − 2 + e−10x
+1
4
e10x + 2 + e−10x
=
4
2
5x
e + e−5x
=
2
1 + (y 0 )2 =
Thus the arc length is given by
Z 3p
Z
1 3 5x
0
2
L =
1 + (y ) dx =
e + e−5x dx
2 0
0
3
1 5x
e − e−5x =
10
0
1 15
=
e − e−15
10
Z
2. The surface area is given by S = 2π
10
9x3
p
1 + (27x2 )2 dx Use the substitution u =
0
1 + 729x4 and du = 4(729)x3 to rewrite the surface area as
Z 10
√
9x3 1 + 729x4 dx
S = 2π
0
7,290,001
Z 7,290,001 1/2
u
π 3/2 = 2π
du =
u 324
243
1
1
π
3/2
=
(7290001) − 1
243
3.
Orient the xy-axes so that the base of the trapezoid
is along the x-axis. The width
2 − 1.5
of the trapezoid at height y is then L(y) = 1.5 +
y, and at that height the depth of
1.8
the water is 1.4 − y. The work required to empty the trough is given by
Z 1.4
2 − 1.5
F = (9.8)1000
(1.4 − y) 1.5 +
y (5) dy
1.8
0
"
3 # 1.4
2−1.5
2
[(1.4) 2−1.5
−
1.5]y
y 1.8
= (9.8)(1000)(5) (1.4)(1.5)y +
− 1.8
2
3
0
= 97462.815 Newtons
4. The curves intersect when −x2 + 4x + 2 = x + 2, or when −x2 + 3x = 0, so when x = 0
and x = 3. Then
Z 3
9
(−x2 + 3x) dx = ρ
M =ρ
2
0
and
Z
3
2
(−x + 3x)
Mx = ρ
0
Therefore, x̄ =
5.
−x2 + 5x + 4
2
99
dx = ρ
5
Z
My = ρ
3
x(−x2 + 3x) dx =
0
27
3
Mx
99
22
My
=
· 29 = and ȳ =
=
· 29 = .
M
4
2
M
5
9
(a) The fluid force, in pounds, on one end of the trough is given by
Z 1
F = 50
2y 1/4 (1 − y) dy
Z0 1 1
5
= 100
y 4 − y 4 dy
0
1
4 5 4 9 4
4
= 100
y − y 5
9
0
1600
100 · 16
=
≈ 35.556 pounds
=
45
45
(b) The work required to empty the swill is given by
Z 1
W = 50
2y 1/4 · 10 · (1 − y) dy
0
Z 1
1
5
4
4
y −y
= 1000
dy
0
1
4 5 4 9 = 1000
y4 − y4 5
9
0
1000 · 16
=
≈ 355.556 foot pounds
45
27
ρ
4
6. Use long-division to rewrite the integral as
Z
Z 18x3 + 5x2 + 128x + 35
2x
18x + 5 + 2
dx =
dx
x2 + 7
x +7
= 9x2 + 5x + ln(x2 + 5) + C
Z
7. For
ln x dx, use dv = dx and u = ln x to obtain
Z
Z
ln x dx = x ln x −
Z
For
1 dx = x ln x − x + C
xn ln x dx when n 6= −1, use dv = xn dx and u = ln x to obtain
Z
xn+1 ln x
−
x ln x dx =
n+1
n
Z
xn
xn+1 ln x
xn+1
dx =
−
+C
n+1
n+1
(n + 1)2
Notice when n = 0, this agrees with the first case. In the case n = −1, use the straight
u-substitution: u = ln x, du = x−1 dx, to get
Z
(ln x)2
ln x
dx =
+C
x
2
8. We use x3 to start the u-column and e−5x to start the dv-column
u
x3
(+)
dv
e−5x
&
(−)
3x
2
&
(+)
6x
&
(−)
&
e−5x
125
e−5x
625
0
Thus
e−5x
25
−
6
Z
e−5x
−
5
3 −5x
xe
x3 e−5x 3x2 e−5x 6xe−5x 6e−5x
dx = −
−
−
−
+ C.
5
25
125
625
Z
π/5
9. For the integral
sin(2x) sin x dx we use the identity sin(2x) = 2 sin x cos x and then
0
π/5
Z
π/5
Z
2 sin2 x cos x dx
sin(2x) sin x dx =
0
0
Z
π/5
sin2 x cos x dx
0
π/5
sin3 x = 2
3 0
2 3 π sin
=
3
5
= 2
Z
10. For the integral
√
x2
Then the integral becomes
Z
1
dx we use the substitution x = 11 sin t and dx = 11 cos t dt.
121 − x2
1
√
dx =
x2 121 − x2
Z
11 cos t dt
121 sin2 t · 11 cos t
Z
1
=
csc2 t dt
121
1
cot t + C
= −
121
√
121 − x2
= −
121x
11. (a) This is a u2 − a2 form so we use u = a sec t and so
2
2
(b) This is a u + a form so we use u = a tan t and so
√
√
8x = 2 sec t and so x =
5x =
√
2 tan t and so x =
√1
2
q
2
5
sec t
tan t
(c) Completing the square the term in the square root becomes 7(x + 3)2 − 3 and
q so this of
√
√
2
2
the form u − a and we use u = a sec t and so 7(x + 3) = 3 sec t and then x = 37 sec t − 3
2
(d) The expression under the radical is
so this has the form a2 − u2 and so
q4 − 6(x + 3) and q
we use u = a sin t and thus (x + 3) = 23 sin t and so x = 23 sin t − 3