D-ARCH
Mathematics
Fall 2015
Solutions – Week 2
Integration
1. Compute the following integrals.
ˆ
ˆ
3
(a)
(x + 4x) dx
(b)
e−4x dx
(c)
ˆ
ˆ
(d)
cos(5x − 2) dx (e)
cosh(5x − 2) dx (f)
ˆ 5
ˆ 65
1
1
(g)
dx
(h)
dx
3x − 3
1 x+3
9
ˆ √
3x dx
ˆ
(2x − 5)−3/2 dx
Solutions (plus a constant for (a)-(f)) :
(a) x4 /4 + 2x2
(b)
− e−4x
(d)
sin (5x − 2) /5
(e)
sinh (5x − 2) /5
(g)
ln 2
(h)
ln 2
√
(c) 2x3/2 / 3
√
(f) − 1/ 2x − 5
2. Compute the following integrals. (These are a bit more advanced!)
ˆ
ˆ
ˆ
1
1
√
(a)
x ln x dx
(b)
dx
dx
(c)
2
2
x (x − 1)
x2 x2 + 1
ˆ − √x
ˆ
ˆ
e
dx
x dx
√ dx
(d)
(e)
dx
(f)
dx
2
x ln x
(x − 1)2/3
x
Solutions (all plus a constant) :
(a)
1 2
x (2 ln |x| − 1) (b)
4
(d)
− 2e−
√
x
(e)
−
√
x2 + 1/x (c)
log log x
(f)
1 1
1
+ ln |x − 1| − log |x + 1|
x 2
2
1/3
3 2
x −1
2
For (a), integrate by parts. For (b), consider the substitution x = tan u (so
√
that x2 + 1 = 1/ cos u). For (c), decompose into partial fractions. For
the others, identify a function g (x) such that the integrand is g 0 (x) times
some function of g (x), so you can use the chain rule “in reverse” (or make
the substitution u = g (x), if you prefer). For example, in (f) we can take
g (x) = x2 .
1
3. Figure 1 shows the graphs of the functions
f (x) = 4x3 + 2x2 − 5x − 2
g(x) = 2x2 − x − 2.
and
10
8
6
4
2
0
−2
−4
−6
f(x)
g(x)
−8
−10
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Figure 1: The functions f (x) and g(x) of Exercise 2
(a) Determine the x-coordinates x1 < x2 < x3 of the points where the
graphs intersect. Solution : x1 = −1, x2 = 0, x3 = 1.
ˆ x3
(b) Calculate the integral
f (x) − g(x) dx. Solution : 0
x1
(c) Calculate the shaded area. Solution : Where f is above g, integrate
f − g. Otherwise, integrate g − f . So the area is
ˆ 1
ˆ 0
(g (x) − f (x)) dx = 2.
(f (x) − g (x)) dx +
−1
0
4. For which x ∈ (0, 3π/2) is f (x) =
´ 2x
x
sin t
dt
t
a local maximum?
Solution : The fundamental theorem of calculus says that
ˆ x
sin t
sin x
d
dt =
,
dx 0 t
x
and we can write
ˆ
f (x) =
0
2x
sin t
dt −
t
ˆ
0
x
sin t
dt,
t
so the chain rule says
f 0 (x) = 2
2
sin 2x sin x
−
.
2x
x
This is zero when sin 2x = sin x. The double-angle formula gives sin 2x =
2 sin x cos x, so sin 2x = sin x occurs when sin x = 0 or when cos x = 1/2. (in
our range, this is when x = π or x = π/3). These are the stationary points
of f (x), where local maxima could occur. Then, use the second-derivative
test to see that only x = π/3 gives a local maximum.
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