Math 103
The University of British Columbia
Solutions, Midterm 2 - March 16, 2011
Mathematics 103: Integral Calculus with Applications to Life Sciences
1. (10 points) Short Answer Problems (show your work).
a. (3)
4p
Z
1 + (f 0 (x))2 dx,
L=
with
f 0 (x) =
p
x3 − 1
1
Z
4
L=
1
√
62
2 5 4 2
x dx = x 2 = (4 · 4 · 4 − 1) =
5
5
5
1
3
2
b. (4)
1
1
1
pred = , pblue = , pgreen =
2
3
6
P (X ≥ 2) = P (X = 2) + P (X = 3) + P (X = 4) = 1 − P (X = 0) − P (X = 1) =
4
1 2 3 33
11
4 0
4 1
2
4
3
−4
=
=
1−
pblue (1 − pblue ) −
pblue (1 − pblue ) = 1 −
3
3 3
81
27
0
1
c. (3)
Z
I=
1
dx =
2
x + 2x + 2
Z
1
dx = arctan(x + 1) + C
(x + 1)2 + 1
2. (10 points) Evaluate the integrals:
a. (4)
3π
2
Z
I1 =
cos(t)e2 sin(t) dt
0
Substitution u = 2 sin(t), dt =
du
2 cos(t) :
1
2
Z
eu du =
eu
+C
2
Substitute back or change boundaries (x = 0 → u = 0, x = 3π
2 → u = −2):
3π
eu −2 1 1
e2 sin(t) 2
I1 =
= =
−1
2 0
2 0
2 e2
b. (6)
Z
I2 =
x−1
dx =
2
x −x−2
Partial fractions:
Z
x−1
dx
(x − 2)(x + 1)
Z
A
B
+
dx
x−2 x+1
Solve A(x + 1) + B(x − 2) = x − 1 for any x. Hence solve A + B = 1 and A − 2B = −1
or evaluate at x = 2 and x = −1: A = 31 , B = 23 .
Z
Z
1
1
2
1
1
2
I2 =
dx +
dx = ln |x − 2| + ln |x + 1| + C
3
x−2
3
x+1
3
3
I2 =
Midterm 2
Page 1 of 3
Math 103
3. (9 points)
a. (3)
Z
M=
1
2
Z
ρ(x)dx =
0
1
2
0
1
2
1
1
cos(πx)dx = sin(πx) = [kg]
π
π
0
b. (2)
ρ̄ =
1
2
M
2
= [kg/m]
π
−0
c. (4)
x̄ =
1
M
Z
1
2
ρ(x)xdx
0
Integration by parts: u = x, dv = ρ(x)
1
x̄ =
M
Z
0
1
2
1
Z 1
2
2
1
1
cos(πx)xdx =
sin(πx)dx =
x sin(πx) −
πM
πM 0
0
1
πM
1
2
1 1
1
x sin(πx) + cos(πx) = − [m]
π
2 π
0
4. (9 points) Part I:
a. (5)
1
0
1
2
3
4
5
cumulative probability
probability P(X=k)
0.8
0.6
0.4
0.2
0
(i)
1
0
0.1
0.2
1
2
3
4
1
2
3
4
5
0.7
0.9
1
3
4
5
0.8
0.6
0.4
0.2
0.2
number of correct answers k
0
0
5
(ii)
0.1
0
1
2
number of correct answers k
b. (2)
P (X ≤ 4) = 1 − P (X = 5) = 0.9
Part II: (2) Expected winnings:
x̄ = 0.1 · $5 + 0.02 · $100 + 0.001 · $500 + 0.0001 · $10, 000 = $0.5 + $2 + $0.5 + $1 = $4
Taking the costs of the ticket of $10 into account, the expected winnings are actually a loss
with an expected net gain of −$6 (both conclusions acceptable).
Midterm 2
Page 2 of 3
Math 103
5. (12 points)
a. (8)
(i)
(ii)
(iii)
(iv)
Solve by rotating part of a circle around x-axis.
√
Circle with radius R: y = f (x) = R2 − x2 .
√
√
√
Find integration boundaries: a = − 22 − 12 = − 3, b = 3.
√
√
Volume of drill hole, radius r: 2 3r2 π = 2 3π.
√
Rb
(v) Thus, V = a πf (x)2 dx − 2 3.
(vi)
√
Z
√
3
√ R
− 3
2
− x2 dx = 2
Z
0
3
√3
√
√
√
1
R2 − x2 dx = 2 R2 x − x3 = 2(4 3 − 3) = 6 3π
3
0
√
(vii) Hence V = 4 3π [cm3 ].
Alternative solution using the shell-method and substitution u = R2 − y 2 :
Z 2
p
V =
2πy(2 R2 − y 2 )dy
1
b. (4*)
(i) Use integration boundaries a, b from above.
√
Z
A=2
3
√
√
p
R2
−
x2 −1
Z
dx = 4
− 3
(ii) Trigonometric substitution
3p
R2
−
√
x2 dx−4
0
√
Z
3
r
1−
3 = 4R
0
x 2
R
√
dx−4 3
x
R
= sin(u), dx = R cos(u)du.
√
(iii) Change limits: x = 0 → u = 0, x = 3 → u = π3
(iv)
Z √3r
Z π
Z πp
x 2
3
3
2
1−
1 − sin(u) R cos(u)du = R
cos(u)2 du =
dx =
R
0
0
0
!
√
π
Z π
3
R 3
R
3
R 1
π
cos(2u) + 1 du =
sin(2u) + u =
+
2 0
2 2
2
4
3
0
√
(v) Insert R = 2; multiply by 4R (see (ii) above); subtract 4 3:
A=
√
8π
− 2 3 [cm2 ]
3
Alternatively, this question can be solved using geometry.
Midterm 2
Page 3 of 3