Solutions to Homework Assignment #3
1. [4 marks] Evaluate the following integrals:
Z
Z
x x
(a)
f (e )e dx, where
f (u) du = sin u.
Z
√
Z
2
3
(b)
5
4
6
(2x + 3x )f (x + x ) dx, where
Z
0
Z
π/4
2
(c)
f (u) du = u2 .
f (tan x)(sec x) dx where
0
Z 2
(d)
0
f (u) du = eu .
π
f (x) + xf (x) +
f
(x + 1)2
0
1
x+1
Z
dx, where
f (u) du =
√
1 + u2 , f (2) = λ.
Solution:
(a) We make the substitution u = ex :
Z
Z
x x
f (e )e dx = f (u) du = sin u + C = sin(ex ) + C.
(b) We make the substitution u = x4 + x6 :
Z
√
2
1
(2x + 3x )f (x + x ) dx =
2
3
0
5
4
Z
12
6
0
1 12 e12 − 1
.
f (u) du = eu =
0
2
2
(c) We make the substitution u = tan x:
Z
Z
π/4
2
f (tan x)(sec x) dx =
0
0
1
1
f (u) du = u2 = 1.
0
(d)
1
π
f
f (x) + xf (x) +
dx =
(x + 1)2
x+1
0
Z 2
Z 2
d
π
1
dx =
f
(xf (x)) dx +
2
x+1
0 dx
0 (x + 1)
Z 1/3
2
xf (x) −π
f (u) du (u = (x + 1)−1 , du = −(x + 1)−2 dx)
0
1
!
√
1/3
√
√
10
− 2 .
= 2λ − π 1 + u2 = 2λ − π
3
1
Z 2
0
1
2. [4 marks] Evaluate the following integrals:
Z
(a) xf (x) dx where f (x) = ln x.
Z
3
(b)
Z
−3
x7 − x5 + 3x
√
dx
x4 + x2 + 1
(sin x)3 (cos x)3 dx
(c)
Z
(sec x)3 dx
(d)
Solution:
(a)
Z
Z
x2
xf (x) dx =
x ln x dx =
ln x −
2
x2
x2
=
ln x −
+ C.
2
4
Z
x2
1
× dx
2
x
x7 − x5 + 3x
is an odd function and therefore
(b) By inspection √
x4 + x2 + 1
Z
a
f (x) dx = 0 for any a.
−a
(c)
Z
Z
3
3
Z
(sin x) (cos x) sin x dx = (1 − (cos x)2 )(cos x)3 sin x dx
Z
Z
3
=
(cos x) sin x dx − (cos x)5 sin x dx
(sin x) (cos x) dx =
= −
2
3
(cos x)4 (cos x)6
+
+ C.
4
6
(d)
Z
Z
d
(sec x) dx =
sec x (tan x) dx = sec x tan x − tan x sec x tan x dx
dx
Z
= sec x tan x − sec x(sec2 x − 1) dx (sec2 x = tan2 x + 1)
Z
Z
= sec x tan x + sec x dx − (sec x)3 dx =⇒
Z
3
Z
(sec x)3 dx =
sec x tan x + ln | sec x + tan x|
+ C.
2
2
3. [4 marks]
(a) Find the area bounded by y = cos 2πx and y = 4x2 − 1/4.
(b) Find the area bounded by y = −|x| + 2 and y = x2 − 4.
ln x
, above the x-axis and to the right of x = 1.
x2
x+1
(d) Find the volume of the solid you get by revolving y = √ , 1 ≤ x ≤ e about the x-axis.
x
(c) Find the area of the region below y =
Solution:
(a) By inspection the graphs intersect at x = ±1/4. Using the obvious symmetry the area is
Z
Z
cos 2πx − (4x − 1/4) dx = 2
cos 2πx − (4x2 − 1/4) dx
−1/4
0
3
sin 2πx 1/4
x 1/4 1
1
x
= 2
−2 4 −
= +
2π
3
4 0
π 12.
0
1/4
2
A =
1/4
(b) The graphs intersect at x = ±2 and therefore the area is
Z 2
Z 2
2
A = 2
(−|x| + 2 − (x − 4)) dx (using symmetry) = 2
(−x + 2 − (x2 − 4)) dx
0
0
Z 2
8
44
= 2
(6 − x − x2 ) dx = 2 12 − 2 −
= .
3
3
0
(c) The area is given by
Z R
Z R
d
ln x
−1
A = lim
(−x ) ln x dx
dx = lim
R→∞ 1
R→∞ 1
x2
dx
R R Z R
ln R
−1
−2
−1 = lim −x ln x +
−x = 1.
x dx = lim −
R→∞
R→∞
R
1
1
1
(d) The volume of the solid you get by rotating y = f (x), a ≤ x ≤ b around the x-axis is
Z b
V =π
(f (x))2 dx. Therefore, in this case
a
2
Z e
e 1
x2 + 2x + 1
e −1
x+2+
= π
dx = π
dx = π
+ 2(e − 1) + ln x 1
x
x
2
1 2
1
2
e −1
e −1
= π
+ 2(e − 1) + 1 = π
+ 2e − 1 .
2
2
Z
V
e
3
4. [4 marks]
Z
(a) Determine
2
Z
2 00
0
2
x f (x) dx, where f (2) = a, f (2) = b and
0
Z
f (x) dx = c.
0
t2
(b) Find F (t) if F (t) =
e−x dx.
0
Z
1
(c) Determine
dx.
2
x +9
Z
1
(d) Determine
dx.
2
x + 4x + 5
Solution:
00
2
(a) We apply the integration by parts formula twice:
Z 2
Z 2
2 Z 2
2
2 00
2 0
0
x f (x) dx = x f (x) −
2xf (x) dx = 4a − 2xf (x) −2
f (x) dx
0
0
= 4a − 4b + 2c.
0
0
Z
0
(b) One version of the FTC is that G (t) = g(t), where G(t) =
0
t
g(x) dx. Changing the
a
upper limit of integration
to another function of t, say α(t), and using the chain rule, gives
!
Z α(t)
d
g(x) dx = g(α(t))α0(t). In this case we get
dt
a
F 0 (t) = 2te−t , and therefore F 00 (t) = (2 − 8t4 )e−t .
4
2
(c)
Z
Z
(d)
1
dx =
2
x + 4x + 5
Z
Z
1
1
1
dx = =
dx
2
x +9
9
(x/3)2 + 1
1
x
=
arctan + C.
3
3
1
dx = arctan(x + 2) + C.
(x + 2)2 + 1
5. [4 marks]
In each of the cases below find n such that the indicated rule will calculate
indicated
accuracy:
Z
3
(a)
e−x dx, trapezoidal rule to within 5 × 10−6 .
2
1
4
Rb
a
f (x)dx to the
Z
3
(b)
Z
e−x dx, midpoint rule to within 5 × 10−8 .
2
1
3
(c)
e−x dx, Simpson’s rule to within 5 × 10−4 .
2
1
Z
(d)
3
e−x dx, Simpson’s rule to within 5 × 10−8 .
2
1
Solution:
We use the error estimates for each of the rules. Let
K = maxa≤x≤b |f 00 (x)|, L = maxa≤x≤b |f (4) (x)|.
Then
Z
K(b − a)3
12n2
a
Z b
K(b − a)3
f (x)dx = Mn + EM(n), |EM(n)| ≤
24n2
a
Z b
L(b − a)5
f (x)dx = Sn + ES(n), |ES(n)| ≤
180n4
a
b
f (x)dx = Tn + ET (n), |ET (n)| ≤
In this case f (x) = e−x and so f 00 (x)p
= (4x2 − 2)e−x and f (4) (x) = (12 − 48x2 + 16x4 )e−x .
In class it was shown that K = f 00 ( 3/2) = 4e−3/2 ≈ 0.8925206404 and L = |f (4) (1)| =
20/e ≈ 7.357588824.
2
2
2
(a) We must find n such that |ET (n)| ≤ 5 × 10−6 . Therefore it suffices to find n satisfying
0.8925206404 × 8
≤ 5 × 10−6 . By calculation we see that n = 355 will do.
12n2
(b) We must find n such that |EM(n)| ≤ 5 × 10−8. Therefore it suffices to find n satisfying
0.8925206404 × 8
≤ 5 × 10−8 . By calculation we see that n = 2440 will do.
24n2
(c) We must find n such that |ES(n)| ≤ 5 × 10−4 . Therefore it suffices to find n satisfying
7.357588824 × 32
≤ 5 × 10−4 . By calculation we see that n = 8 will do.
4
180n
(d) We must find n such that |ES(n)| ≤ 5 × 10−8 . Therefore it suffices to find n satisfying
7.357588824 × 32
≤ 5 × 10−8. By calculation we see that n = 72 will do.
180n4
5