Marks
1. Short-Answer Questions. Put your answers in the boxes provided but show your work
also. Each question is worth 3 marks, but not all questions are of equal difficulty. At most
one mark will be given for an incorrect answer. Unless otherwise stated, simplify your
answers as much as possible.
Z
[3]
(a) Evaluate
3 + x5
√ dx
x
Z
3 + x5
√
dx =
x
Z
√
2
3x−1/2 + x9/2 dx = 6 x + x11/2 + C.
11
n
X
π
iπ
tan
n→∞
4n
4n
(b) What integral is defined by the following expression? lim
i=1
[3]
Do not evaluate the integral.
If one divides [0, π/4] into n equal subintervals, then ∆x =
Therefore,
π
4n
and xi = i∆x =
Z π/4
n
n
X
X
iπ
π
tan x dx.
tan
= lim
tan(xi )∆x =
lim
n→∞
4n
4n n→∞
0
i=1
i=1
1
iπ
4n .
April 24, 2009
1
Z
[3]
(c) Evaluate
Math 101
Page 2 of 13 pages
p
(y + 1) 2y + y 2 dy
0
Let u = 2y + y 2 . Then du = (2 + 2y) dy = 2(y + 1) dy. When y = 0, u = 0 and when
y = 1, u = 3. Thus,
Z
0
Z
[3]
(d) Evaluate
1
Z
p
1 3√
1 3/2 3 1 3/2 √
2
u
u du =
(y + 1) 2y + y dy =
= 3 = 3.
2 0
3
3
0
x2 ln x dx
Integrate by parts: let u = ln x and dv = x2 dx so that du = dx/x and v = x3 /3.
Therefore,
Z 2
Z
x3
x3
x
x3
2
ln x −
dx =
ln x −
+ C.
x ln x dx =
3
3
3
9
2
April 24, 2009
Math 101
Page 3 of 13 pages
(e) Find the length of the curve x = 100 +
2y 3/2 ,
0 ≤ y ≤ 11.
[3]
dx
dy
√
= 3 y so the length of the curve L is
Z
L=
0
[3]
11 p
1 + 9y dy =
i11
2
2 h
= (1000 − 1) = 74.
(1 + 9y)3/2
27
27
0
(f) What is the average value of | sin θ − cos θ| over the interval 0 ≤ θ ≤ π/2?
Notice that cos θ ≥ sin θ for 0 ≤ θ ≤ π/4 and sin θ ≥ cos θ for π/4 ≤ θ ≤ π/2. Therefore
the average value is
!
Z
Z π/4
Z π/2
2 π/2
2
| sin θ − cos θ| dθ =
(cos θ − sin θ) dθ +
(sin θ − cos θ) dθ
π 0
π
0
π/4
2
π/4
π/2
[sin θ + cos θ]0 + [− cos θ − sin θ]π/4
=
π
√ 2√
( 2 − 1) + (−1 + 2)
=
π√
4( 2 − 1)
.
=
π
3
April 24, 2009
Math 101
Page 4 of 13 pages
(g) Give the first three nonzero terms of the Maclaurin series (power series in x) for
[3]
Z
2
e−x − 1
dx
x
Recall that
ex = 1 + x +
x2 x3
+
+ ...
2
6
Therefore,
2
e−x = 1 − x2 +
and
x4 x6
−
+ ...,
2
6
2
x3 x5
e−x − 1
= −x +
−
+ ....
x
2
6
Finally,
Z
[3]
2
e−x − 1
x2 x4 x6
dx = − +
−
+ ...
x
2
8
36
(h) For what values of r does the function y = xr satisfy the following differential equation
(for x > 0)?
x2 y 00 + 4xy 0 + 2y = 0
If y = xr then y 0 = rxr−1 and y 00 = r(r − 1)xr−2 . Therefore,
x2 y 00 + 4xy 0 + 2y = r(r − 1)xr + 4rxr + 2xr = (r2 + 3r + 2)xr = (r + 1)(r + 2)xr .
This will be zero when r = −1 or r = −2.
4
April 24, 2009
[3]
Math 101
Page 5 of 13 pages
(i) Calculate the derivative f 0 (x) if f (x) =
ex
Z
√
cos t dt.
x
Z
ex
√
x
Z
cos t dt =
ex
√
Z
cos t dt −
0
x√
cos t dt.
0
Therefore, by the chain rule and the first fundamental theorem of calculus,
√
√
f 0 (x) = ex cos ex − cos x.
Z
(j) Give the Simpson’s rule approximation to
2
sin(ex ) dx using 4 equal subintervals (do
0
[3]
not simplify your answer).
Here ∆x = 1/2, x0 = 0, x1 = 1/2, x2 = 1, x3 = 3/2, x4 = 2. Therefore,
S4 =
1
sin(1) + 4 sin(e1/2 ) + 2 sin(e) + 4 sin(e3/2 ) + sin(e2 ) .
6
5
April 24, 2009
Math 101
Page 6 of 13 pages
Full-Solution Problems. In the remaining questions, justify your answers and show all your
work. If a box is provided, write your final answer there. If you need more space, use the
back of the previous page. Unless otherwise indicated, simplification of answers is
not required.
∞
Z
[6]
2. For what values of p does
e
dx
converge?
x(ln x)p
Z
∞
e
dx
= lim
x(ln x)p t→∞
Z
t
e
dx
.
x(ln x)p
Let u = ln x so that du = dx/x. When x = e, u = 1 and when x = t, u = ln t. Therefore,
t
Z
lim
t→∞ e
dx
= lim
x(ln x)p t→∞
Z
ln t
1
du
.
up
If p = 1, then
Z
ln t
lim
t→∞ 1
Z
du
up
=
=
=
ln t
lim
t→∞ 1
du
u
t
lim [ln u]ln
1
t→∞
lim ln(ln t) = ∞.
t→∞
Therefore if p = 1, the integral does not converge.
If p 6= 1, then
Z
lim
t→∞ 1
ln t
du
up
=
lim
t→∞
u1−p
1−p
ln t
1
(ln t)1−p
1
= lim
+
t→∞ 1 − p
p−1
∞
if p < 1
=
1
p−1 if p > 1
Therefore, the integral converges if and only if p > 1.
6
April 24, 2009
[10]
Math 101
Page 7 of 13 pages
3. Let R be the finite region in the xy-plane bounded by x = 0, y = 0 and y = cos x.
(a) Calculate the centroid of R.
The area A of R is
Z
π/2
A=
π/2
cos x dx = [sin x]0
0
= 1.
The x-coordinate of the centroid x is
Z π/2
π
π/2
x=
x cos x dx = [x sin x + cos x]0 = − 1.
2
0
We used integration by parts to evaluate the integral.
The y-coordinate of the centroid y is
y =
=
=
=
1
2
π/2
Z
cos2 x dx
Z
1 π/2 1 cos 2x
dx
+
2 0
2
2
1
sin 2x π/2
x+
4
2
0
π
.
8
0
Therefore, the centroid of R is (π/2 − 1, π/8).
(b) Express the volume of the solid obtained by rotating R about the line x = −2 as an
integral. Do not evaluate the integral.
Z
V = 2π
π/2
(x + 2) cos x dx.
0
7
April 24, 2009
[6]
[6]
Math 101
Page 8 of 13 pages
4. Calculate the integrals:
Z
(a) (4 − x2 )−3/2 dx
Let x = 2 sin u so that dx = 2 cos u du and 4 − x2 = 4 cos2 u. Thus,
Z
Z
2 cos u du
2 −3/2
(4 − x )
dx =
8 cos3 u
Z
1
sec2 u du
=
4
1
=
tan u + C
4
x
√
+ C.
=
4 4 − x2
Z
√
(b)
1 + ex dx
Let u =
√
1 + ex , i.e. u2 = 1 + ex . Then 2u du = ex dx = (u2 − 1) dx. Therefore,
Z
Z
Z
√
2u2
2
x
1 + e dx =
du = (2 + 2
) du.
2
u −1
u −1
By a partial fraction decomposition,
u2
2
1
1
2
=
=
−
.
−1
(u − 1)(u + 1)
u−1 u+1
Thus,
Z
Z
Z
2
du
du
(2 + 2
) du = 2u +
−
u −1
u−1
u+1
= 2u + ln |u − 1| − ln |u + 1| + C
u − 1
+C
= 2u + ln u + 1
√
1 + ex − 1 √
x
+ C.
= 2 1 + e + ln √
1 + ex + 1 8
April 24, 2009
[10]
Math 101
Page 9 of 13 pages
5. Let X be a random variable with probability density function
f (x) =
kx(1 − x4 ) if 0 ≤ x ≤ 1,
0
if x < 0 or x > 1.
(a) Find the value of k.
1
1
x2 x6
−
= .
x(1 − x ) dx =
2
6
3
0
0
R1
Therefore, we need to choose k = 3 in order for 0 f (x) dx = 1.
Z
1
4
(b) Find the mean µ.
Z
µ=
1
Z
xf (x) dx = 3
0
0
1
x3 x7
−
x (1 − x ) dx = 3
3
7
2
4
1
0
4
= .
7
(c) Find an algebraic equation satisfied by the median m.
The mean m is such that
1
=3
2
Z
0
m
x2 x6
−
x(1 − x ) dx = 3
2
6
4
or
m6 − 3m2 + 1 = 0.
9
m
0
3
m6
= m2 −
2
2
April 24, 2009
Math 101
Page 10 of 13 pages
6. Solve these differential equations:
[6]
(a) y 0 = xy 2 with initial conditions y(0) = 1.
This is a separable equation:
dy
dx
dy
y2
Z
dy
y2
1
−
y
= xy 2
= x dx
Z
=
x dx
=
x2
+ C.
2
The initial condition y(0) = 1 implies that C = −1. Therefore,
y=
[6]
1
.
1 − x2 /2
(b) y 0 − 2y = 4 + e3t . (Give the general solution)
R
This is a linear equation. The integrating factor is I(t) = e
both sides of the equation by I(t) yields
−2 dt
e−2t y 0 − 2e−2t y = 4e−2t + et
(e−2t y)0 = 4e−2t + et
Z
−2t
e y =
(4e−2t + et ) dt
e−2t y = −2e−2t + et + C
y = −2 + e3t + Ce2t .
10
= e−2t . Multiplying
April 24, 2009
[10]
Math 101
Page 11 of 13 pages
7. An open metal tank has two ends which are isosceles triangles with vertex at the bottom,
two sides which are rectangular, and an open top. The tank is 1 metre wide, 2 metres deep.
10 metres long and full of water (density = 1000kg/m3 )
(a) What is the hydrostatic force on each triangular end?
We choose a vertical x-axis with origin at the bottom of the tank. We divide the interval
[0, 2] into n equal subintervals of length ∆x. The i-th horizontal strip is approximately
a rectangle of height ∆x and width wi where by similar triangles
1
wi
=
xi
2
or wi =
xi
.
2
Therefore the area of the i-th strip is approximately xi /2 ∆x and the hydrostatic force
on the i-th strip is
Fi ≈ 9800(2 − xi )xi /2 ∆x.
Therefore the total hydrostatic force is
F
=
lim
n→∞
n
X
4900(2 − xi )xi ∆x
i=1
Z 2
(2 − x)x dx
= 4900
0
2
x3
2
= 4900 x −
3 0
19600
=
N.
3
(b) Express the work required to pump the water out of the tank, if it is pumped out over
the top edge, as an integral. (Do not evaluate the integral.)
With the same setup as in (a), the volume of the i-th slab is 10 · xi /2 · ∆x = 5xi ∆x.
Therefore the work Wi to pump the i-th slab is 9800 · 5(2 − xi )xi ∆x and the total work
done is
Z 2
W = 49000
(2 − x)x dx.
0
(c) Express the hydrostatic force on each of the rectangular sides as an integral. (Do not
evaluate the integral.)
√
Note that the i-th strip is a rectangle with height ( 17/4)∆x and width 10. Therefore
the force is
√ Z 2
2450 17
x dx.
0
11
April 24, 2009
[10]
Math 101
Page 12 of 13 pages
8. Consider the chemical reaction
A + B → C + D.
Suppose at time t = 0 sec the concentration of chemical A is 0.1 mol/L, the concentration
of chemical B is 0.2 mol/L, and the concentrations of chemicals C and D are both 0. For
t ≥ 0, let x(t) be the concentration of chemical D in mol/L. It can be shown that x(t) is the
solution to the initial-value problem
dx
= k(0.1 − x)(0.2 − x),
dt
x(0) = 0,
where k is a positive constant whose value can be determined by experiment.
(a) Solve the initial-value problem to find x(t) explicitly.
This is a separable equation.
dx
(0.1 − x)(0.2 − x)
Z
dx
(0.1 − x)(0.2 − x)
= k dt
Z
=
k dt
To evaluate the left hand side, we use partial fractions:
1
10
10
=
−
,
(0.1 − x)(0.2 − x)
0.1 − x 0.2 − x
and therefore,
Z
10
10
dx =
k dt
−
0.1 − x 0.2 − x
ln |0.2 − x| − ln |0.1 − x| = kt + C
0.2 − x = kt + C
ln 0.1 − x Z 0.2 − x
0.1 − x
= Aekt
The initial condition x(0) = 0 implies that A = 2. Therefore,
0.2 − x
0.1 − x
0.1
1+
0.1 − x
0.1
0.1 − x
= 2ekt
= 2ekt
= 2ekt − 1
0.1 − x =
1
20ekt − 10
x = 0.1 −
1
− 10
20ekt
(b) What value does the concentration of chemical D approach as t approaches ∞?
lim x(t) = 0.1
t→∞
(we could’ve guessed this based on the physical interpretation).
12