MATH 101, Section 212 (CSP)
Week 2: Marked Homework Solutions
2011 Jan 20
1. (a) [1] f 0 (t) = t2 + 2t + 4.
Ru 2
(b) [2] Write y = π e−t dt, where u = sin x. Then by the Chain Rule
Z u
dy du
d
d
dy
2
2
2
=
·
=
sin x = e−u · cos x = e− sin x · cos x.
e−t dt ·
dx
du dx
du π
dx
(c) [3] g(x) =
R0
2x
cos(t2 ) dt +
g 0 (x) = cos((x3 )2 ) ·
R x3
0
cos(t2 ) dt =
R x3
0
cos(t2 ) dt −
R 2x
0
cos(t2 ) dt, then
d
d 3
(x ) − cos((2x)2 ) · (2x) = 3x2 cos(x6 ) − 2 cos(4x2 ).
dx
dx
R x2 −2x 1
Ru
d
d
2. [5] Find the critical point(s): f 0 (x) = dx
dt = du
1+t4
0
0
x2 − 2x, so
2x − 2
1
·
(2x
−
2)
=
.
f 0 (x) =
1 + u4
1 + (x2 − 2x)4
1
1+t4
dt ·
du
dt
where u =
The denominator is always positive, so f 0 (x) has the same sign as the numerator 2x − 2:
f 0 (x) < 0 iff x < 1, f 0 (x) = 0 iff x = 1, f 0 (x) > 0 iff x > 1. So x = 1 is a critical point,
f (x) is decreasing for x < 1 and is increasing for x > 1. By the First Derivative Test, the
absolute minimum value of f is
Z −1
Z 0
1
1
f (1) =
dt
=−
dt .
4
1 + t4
0
−1 1 + t
R3
(t2 + 5t − 1) dt
3
= 31 t3 +
t −1 = 31 (3)3 + 52 (3)2 − (3) − 31 (−1)3 + 52 (−1)2 − (−1) = 76
= 25 13 .
3
√
R2
2
R2√
(b) [2] 1 x dx = 1 x1/2 dx = 32 x3/2 1 = 23 (2)3/2 − 23 (1)3/2 = 23 (2 2 − 1).
Rπ
(c) [2] 0 sin s ds = − cos s]π0 = − cos π − (− cos 0) = 2.
10
R 10
(d) [3] If a 6= 0, then 0 e−at dt = − a1 e−at 0 = − a1 (e−10a − e0 ) = a1 (1 − e−10a ).
R 10
R 10
R 10
If a = 0, then 0 e−at dt = 0 e0 dt = 0 1 dt = 10.
3/2
1
R 3/2
(e) [2] 0 cos πx dx = π1 sin πx 0 = π1 sin 3π
− π sin 0 = − π1 .
2
3. (a) [2]
−1
5 2
t
−
2
4. [3] If N (t) is the population at time t, we are given that dN
= N 0 (t) = 2000 + t3 .
dt
By the Net Change Theorem, the net change in population is
4
Z 4
Z 4
1 4
0
3
N (4) − N (2) =
= 4060.
N (t) dt =
(2000 + t ) dt = 2000t + t
4
2
2
2
1
5. [5] Let s(t) be the position of the particle (in m) at time t (in s), so
s0 (t) = v(t) = cos πt
m/s.
(a) The displacement of the particle along the line during the time interval from t = 0 to
t = 1 is, by the Net Change Theorem,
Z 1
s(1) − s(0) =
cos πt dt = π1 sin πt]10 = π1 (sin π − sin 0) = 0 m.
0
(b) The distance the particle travels along the line during the time interval from t = 0 to
t = 1 is (noting that cos πt changes sign at t = 1/2)
Z 1
Z 1/2
Z 1
| cos πt| dt
| cos πt| dt +
| cos πt| dt =
1/2
0
0
=
=
=
=
Z
1/2
cos πt dt +
0
Z
1
(− cos πt) dt
1/2
1/2
1
1
sin πt 0 − π1 sin πt 1/2
π
1
(1 − 0) − π1 (0 − 1)
π
2
m.
π
(c) The answers to parts (a) and (b) are not the same. In the given time interval, the particle
moves 1/π m in the positive direction, then 1/π m in the negative direction so it ends up
exactly where it started and the displacement is 0 m, but the particle travels a total distance
of (1π) + (1/π) = 2/π m.
6. (a) [2] Substitute u = 4 − 3x, then du = −3 dx (so dx = − 31 du), and
Z
Z
Z
1
1
1 1
1
dx =
(− 3 du) = − 3
du = − 31 ln |u| + C = − 13 ln |4 − 3x| + C.
4 − 3x
u
u
(b) [3] Substitute u = x − 3, then du = dx, and
Z 5
Z 5
Z 2
√
2
1/2
x − 3 dx =
(x − 3) dx =
u1/2 du = 32 u3/2 1 =
4
4
1
Alternatively, evaluate as 32 (x − 3)
3/2 5
4
√
= 32 (2 2 − 1).
2
3
√
23/2 − 13/2 = 23 (2 2 − 1).
(c) [3] Substitute u = t − 1, then du = dt and t = u + 1, and
Z
Z
t2
(u + 1)2
√
√
dt =
du
u
t−1
Z
=
u−1/2 (u2 + 2u + 1) du
Z
= (u3/2 + 2u1/2 + u−1/2 ) du
= 52 u5/2 + 43 u3/2 + 2u1/2 + C
= 25 (t − 1)5/2 + 34 (t − 1)3/2 + 2(t − 1)1/2 + C.
2
(d) [3] An easy way, using symmetry:
Substitute u = x − 1, then du = dx, and
Z 1
Z 2
u
x−1
dx =
du = 0
4
4
−1 1 + u
0 1 + (x − 1)
since the integrand is odd and the limits of integration are symmetric about 0.
Another way, which would still work if the limits of integration in the transformed integral
were not symmetric:
Substitute u = (x − 1)2 , then du = 2(x − 1) dx, and
Z 2
Z 2
x−1
1
dx =
(x − 1) dx
4
2 2
0 1 + (x − 1)
0 1 + [(x − 1) ]
Z 1
1 1
=
du
2 2
1 1+u
= 12 arctan u]11
=
1
2
arctan((x − 1)2 )]20
= 0.
(e) [3] Similar to (c).
Substitute u = 2s + 1, then du = 2 ds and s = (u − 1)/2.
Z 4
Z 9
s
u−1
√
√ du
ds =
2s + 1
0
1 4 u
Z 9
1
= 4
(u1/2 − u−1/2 ) du
1
4
1 2 3/2
= 4 3 u − 2u1/2 1
= 21 31 93/2 − 91/2 − 31 13/2 − 11/2
=
10
.
3
(f) [3] This integral can be done directly R(“guess and Rcheck” the antiderivative), or by
substituting u = ωt, then du = ω dt, and cos ωt dt = cos u ω1 du = ω1 sin ωt + C. Using
symmetry (the integrand is even and the limits of integration are symmetric about 0) saves
a bit of work:
Z π/(2ω)
Z π/(2ω)
π/(2ω)
cos ωt dt = ω2 sin ωt]0
cos ωt dt = 2
= ω2 (sin π2 − sin 0) = ω2 .
0
−π/(2ω)
Without noticing the symmetry, we get the same answer:
Z π/(2ω)
π/(2ω)
cos ωt dt = ω1 sin ωt]−π/(2ω) = ω1 [sin π2 − sin(− π2 )] = ω1 [1 − (−1)] = ω2 .
−π/(2ω)
7. [3] Substitute u = x2 , then du = 2x dx. Note that u = 02 = 0 when x = 0, and u = 32 = 9
when x = 3.
Z
Z 3
Z 3
Z 9
1 9
2
2
1
f (u) du = 12 (4) = 2.
xf (x ) dx =
f (x ) (x dx) =
f (u) 2 du = 2
0
0
0
0
3