MATH 105 – 951
MATH 105 Midterm Practice Problems
Solutions
Short Answer Questions
Evaluate the following integrals or state that they diverge.
Z 5
x
1.
dx
0 x + 10
Solution: Let u = x + 10. Then du = dx and x = u − 10 so that
x
u − 10
dx =
du. When x = 0,
x + 10
u
u = 10, and when x = 5, u = 15. Therefore,
Z 5
Z 15
Z 15 x
u − 10
u 10
dx =
du =
−
du
u
u
u
0 x + 10
10
10
Z 15 10
=
1−
du
u
10
15
= (u − 10 ln |u|) 10
= 15 − 10 ln(15) − 10 + 10 ln(10).
Z
2.
3
4
1
dy
y 2 − 4y − 12
Solution: First, note that y 2 − 4y − 12 = (y − 6)(y + 2). To use the method of partial fractions we need
to find A and B such that
1
A
B
=
+
.
(y − 6)(y + 2)
y−6 y+2
Multiplying both sides by (y − 6)(y + 2) we get that
1 = A(y + 2) + B(y − 6).
Plugging y = −2 into the equation we see that 1 = B · (−8), or B = − 81 . Plugging y = 6 into the
equation we see that 1 = A · (8), or A = 18 . Therefore
1
1/8
−1/8
=
+
,
(y − 6)(y + 2)
y−6 y+2
and so
Z
3
4
Z 4
1/8
−1/8
dy +
dy
3 y−6
3 y+2
4
4
1
1
= ln |y − 6| − ln |y + 2|
8
8
3
3
1
= (ln(2) − ln(3) − ln(6) + ln(5)).
8
1
dy =
2
y − 4y − 12
Z
4
Z
3.
a
√
0
a2
1
dx
− x2
Solution: Note that this is an improper integral, so when evaluating at x = a, we’re really doing
this by evaluating at t√< a and taking the limit of the result as x → a− . Let u = a sin(x). Then
du = a cos(x)dx, and a2 − x2 = a cos(u). Furthermore, when x = 0, sin(u) = 0, so u = 0. When
x = a, sin(u) = 1, so u = π/2. Therefore,
Z
a
0
√
1
dx =
a2 − x2
Z
0
π/2
1
a cos(u)du =
a cos(u)
So, the integral converges, and the value of the integral is
Z
4.
2
∞
Z
0
π/2
π/2
π
= .
du = u 2
0
π
2.
1
dx
x ln(x)
1
1
1
dx and so
dx = du. Furthermore, when x = 2, u = ln(2),
x
x ln(x)
u
and when x = ∞, u = ∞. Therefore,
∞
Z ∞
Z ∞
1
1
= ∞.
dx =
du = ln |u|
x
ln(x)
u
2
ln(2)
ln(2)
Solution: Let u = ln(x). Then du =
Therefore the integral diverges.
Z
5.
tan2 (u) cos2 (u)du
Solution: We have that
Z
Z
Z
sin2 (u)
2
tan2 (u) cos2 (u)du =
cos
(u)du
=
sin2 (u)du
cos2 (u)
Z 1 1
=
− cos(2u) du
2 2
u 1
= − sin(2u) + C
2 4
Z
6.
csc4 (θ) cos(θ)dθ
2
1
. Let u = sin(θ). Then du = cos(θ)dθ, and we get that
sin4 (θ)
Z
Z
1
1
1
csc4 (θ) cos(θ)dθ =
du = − 3 + C = −
+ C.
u4
3u
3 sin3 (θ)
Solution: First recall that csc4 (θ) =
Z
7.
x2 + 2
dx
x+2
Solution: Using long division of polynomials we get that
x2 + 2
6
=x−2+
.
x+2
x+2
So,
Z
Z
8.
x2 + 2
dx =
x+2
Z x−2+
6
x+2
dx =
1 2
x − 2x + 6 ln |x + 2| + C.
2
t cos(t2 )dt
Solution: Let u = t2 . Then du = 2tdt (or 12 du = tdt) and we get that
Z
Z
1
1
1
2
t cos(t )dt =
cos(u)du = sin(u) + C = sin(t2 ) + C.
2
2
2
Z
9.
x3/2 ln(x)dx
Solution: Let u = ln(x) and dv = x3/2 dx so that du = x1 dx and v = 52 x5/2 . Using integration by parts
we have that
Z
Z
2 5/2 1
2
x
dx
x3/2 ln(x) = x5/2 ln(x) −
5
5
x
Z
2
2 3/2
= x5/2 ln(x) −
x dx
5
5
2
4
= x5/2 ln(x) − x5/2 + C.
5
25
3
Z
10.
ln(x)dx
(Hint: Think of ln(x) as 1 · ln(x).)
Solution: Let u = ln(x) and dv = 1dx so that du = x1 dx and v = x. Using integration by parts we have
that
Z
Z
Z
1
ln(x)dx = x ln(x) − x dx = x ln(x) − dx = x ln(x) − x + C.
x
Z
11.
2
√
1
1
dx
x−1
Solution: Note that this is an improper integral, so when we evaluate at x = 1, we’re really doing this
by evaluating at t > 1 and taking the limit of the result as t → 1+ . We have that
2
Z 2
√
1
√
dx = 2 x − 1 = 2.
x−1
1
1
So, the integral converges, and the value of the integral is 2.
Long Answer Questions
Z
1. Let F (x) =
1
x
1
dt. Find the equation of the line tangent to F (x) at x = 2.
t2 + 6t + 5
Solutions: From the FTC we know that F 0 (x) =
x2
2
Z
F (2) =
1
1
, so F 0 (2) =
+ 6x + 5
1
21 .
Also,
1
dt.
t2 + 6t + 5
1
1
=
, so to use the method of partial fractions we need to find A and B
+ 6t + 5
(t + 1)(t + 5)
such that
1
A
B
=
+
.
(t + 1)(t + 5)
t+1 t+5
Now
t2
Multiplying both sides of this equality by (t + 1)(t + 5) we get
1 = A(t + 5) + B(t + 1).
Plugging t = −1 into the above equality gives 1 = A(4), or A = 1/4. Plugging t = −5 into the above
4
equality gives 1 = B(−4), or B = −1/4. Therefore,
2
Z
1
1
dt =
(t + 1)(t + 5)
2
Z
1
=
1/4
−1/4
+
t+1
t+5
dt
2
1
1
ln |t + 1| − ln |t + 5| 4
4
1
1
(ln(3) − ln(7) − ln(2) + ln(6))
4 1
9
= ln
4
7
(It’s not necessary to simplify to get the last equality). So F (2) = 14 ln 79 . Therefore, the equation of
the tangent line is given by
1
9
1
y − ln
=
(x − 2).
4
7
21
=
(You do not need to rearrange the above equation to solve for y).
2. Recall that a differentiable function F (x) is said to be increasing at a point a if F 0 (a) ≥ 0. Show that
the function
Z ex
t
ee dt
0
is always increasing.
x
Z
Solution: Let G(x) =
et
ex
0
Z
e dt. Then, from the FTC, we know that G (x) = e . Now, if F (x) =
0
t
ee dt,
0
then F (x) = G(ex ). So, from the chain rule we have that
F 0 (x) = G0 (ex ) · ex = ee
ex
ex
· ex .
Since ex > 0 for all x, it follows that F 0 (x) > 0 for all x and is therefore always increasing.
Z
3. Find the derivative of the function F (x) =
x
xet dt. (Hint: x does not depend on t.)
0
Z
x
et dt. Then, from the FTC, we have that G0 (x) = ex . Since x does not depend
Z x
Z x
on t (the variable of integration), F (x) =
xet dt = x
et dt = xG(x). Therefore, using the product
Solution: Let G(x) =
0
rule we have that
0
0
d
F (x) =
(xG(x)) = G(x) + xG0 (x) =
dx
0
5
Z
0
x
x
e dt + xe = e + xex = ex − 1 + xex .
t
x
t
0
4. If f (x) is continuous on [a, b], then the average value of f (x) on [a, b] is defined to be
fave =
Find the average value of the function f (x) =
1
b−a
Z
b
f (x)dx.
a
x3 + 4
on the interval [0, 2].
x2 + 4x + 3
Solution: Using long division of polynomials we get that
x2
x3 + 4
13x + 16
=x−4+ 2
.
+ 4x + 3
x + 4x + 3
Now, note that x2 + 4x + 3 = (x + 1)(x + 3). So, to use the method of partial fractions on the second
term above we need to find A and B such that
13x + 16
A
B
=
+
.
(x + 1)(x + 3)
x+1 x+3
Multiplying this through by (x + 1)(x + 3) we get that
13x + 16 = A(x + 3) + B(x + 1).
Plugging x = −3 into the above equation and solving for B we get that B = 23/2. Plugging x = −1
into the above equation and solving for A we get that A = 3/2. So
13x + 16
3/2
23/2
=
+
.
(x + 1)(x + 3)
x+1 x+3
Therefore,
fave
1
=
2
Z
0
2
23/2
3/2
+
x−4+
x+1 x+3
dx
2
3
23
1 1 2
x − 4x + ln |x + 1| +
ln |x + 3| 2 2
2
2
0
1 1 2
3
23
3
23
=
· 2 − 8 + ln(3) +
ln(5) − 0 + 0 − ln(1) −
ln(3)
2 2
2
2
2
2
1
23
=
−6 +
ln(5) − 10 ln(3) .
2
2
=
5. Find the area of the overlapping portion of the circles x2 + (y − 1)2 = 1 and x2 + y 2 = 1 that is in the
first quadrant. (Note: The equation x2 + (y − 1)2 = 1 is the equation of the circle of radius 1 centred
at the point (0, 1), and the equation x2 + y 2 = 1 is the equation of the circle of radius 1 centred at the
origin.)
Solution: First note that the circles intersect when (1 − (y − 1)2 ) + y 2 = 1 (solving x2 + (y − 1)2 = 1
for x2 and then substituting this into the equation of the other circle). So, the circles intersect when
2y − 1 = 0, or, equivalently,
when y = 1/2. Now, if y = 1/2, then substituting this into√x2 + y 2 = 1
√
we see that x = 3/2. Now, the top of the circle centred at the origin is given by y = 1 − x2 , and
6
the bottom of the circle centred at the point (0, 1) is given by y = 1 −
overlapping portion of the circles in the first quadrant is
√
Z
3/2
area =
√
Z
p
p
2
2
[( 1 − x ) − (1 − 1 − x )]dx =
0
3/2
√
1 − x2 . So, the area of the
p
(2 1 − x2 − 1)dx.
0
√
Now, let x = sin(u). √
Then dx = cos(u)du
and cos(u) = 1 − x2 . Also, when x = 0, sin(u) = 0, so
√
u = 0, and when x = 3/2, sin(u) = 3/2, so u = π/3. Using this trig substitution we get
Z
π/3
Z
π/3
(2 cos2 (u) − cos(u))du
(2 cos(u) − 1) cos(u)du =
area =
0
0
Z
π/3
(1 + cos(2u) − cos(u))du
=
0
π/3
1
u + sin(2u) − sin(u) 2
0
√
√
π
3
3
= +
−
−0
3
2
√4
π
3
= −
.
3
4
=
6. Find the area between the curves y = e + sin2 (πx) and y = xex in the first quadrant.
4
3 y
2
1
x
1
Solution: The area between the two curves in the first quadrant is given by the integral
Z 1
Z 1
Z 1
1 1
2
x
area =
[(e + sin (πx)) − xe ]dx =
e + − cos(2πx) dx −
xex dx.
2 2
0
0
0
We can calculate the first integral directly
Z
0
1
1 1
(e + − cos(2πx))dx =
2 2
1
x
1
1
e·x+ −
sin(2πx) = e +
2 4π
2
0
To calculate the second integral we will use the method of integration by parts with u = x and
7
dv = ex dx. Then du = dx, v = ex , and IBP gives us
1
Z
0
1 Z
xex dx = xex −
0
1
ex dx
0
1 !
x
e =e−
0
= e − (e − 1) = 1.
So, the area between the two curves in the first quadrant is area = e − 21 .
7. (a) Use sigma notation to write down a Midpoint Riemann Sum approximation of the area under the
curve y = e1/t between t = 1 and t = 2 with n = 10. Do not evaluate the Riemann sum.
Solution: For this problem a = 1, b = 2, n = 10 and f (t) = e1/t . So, δt =
midpoint is 1 + (2i − 1)/20, and the approximation is given by
2
Z
e1/t dt ≈
1
10
X
e1/(1+
2i−1
20 )
i=1
·
2−1
10
=
1
10 .
So, the ith
1
.
10
(b) Use the error formula
N (b − a)3
24n2
to find a bound for the error in the approximation in (a).
|error| ≤
Solution: (This is a little bit more complicated than I had originally intended...) Here, since we’ve
used the Midpoint Riemann Sum, N represents an upper bound on the second derivative f (t) = e1/t
2
00
1/t
on the interval [1, 2]. Now, if f (t) = e1/t , then f 0 (t) = e1/t · −1
· −1
+ e1/t t23 .
t2 , and so f (t) = e
t2
So on [1, 2],
|e1/t |
|e1/t |
|f 00 (t)| ≤ 4 + 2 3 .
t
t
t
Now, since g(t) = e is an increasing function (the y-values get bigger as t gets bigger) and on
[1, 2],
1
1
1
≤ ≤ ,
2
t
1
we get that |e1/t | ≤ e < 3 for t ∈ [1, 2]. Therefore,
|f 00 (t)| ≤
3
3
+ 2 · = 9 =: N.
1
1
Therefore,
|error| ≤
8
9
.
2400
8. The lifetime T (in hours) of a lightbulb has probability density function
(
1 −t/100
e
if t ≥ 0
f (t) = 100
0
if t < 0
(a) Find the probability that the lightbulb will not burn out within the first 100 hours. The probability
that the lightbulb will not burn out within the first 100 hours is just the probability that the lifetime
of the lightbulb is greater than 100 hours.
∞
Z ∞
Z ∞
1
1 −t/100
−t/100 = .
e
dt = −e
f (t)dt =
P(T > 100) =
e
100 100
100
100
(b) Find the expected lifetime of the lightbulb.
The expected lifetime of the lightbulb is
Z ∞
Z
E(T ) =
tf (t)dt =
−∞
Let u = t and dv =
0
∞
t·
1 −t/100
e
dt.
100
1 −t/100
dt.
100 e
Then du = dt and dv = −e−t/100 and using IBP gives us
∞ Z ∞
−t/100 e−t/100 dt
E(T ) = −te
−
0
0
∞ −t/100 = 0 − −100e
0
= 100.
(where L’Hôpital’s Rule was used to evaluate the limit lim te−t/100 ). Therefore, the expected
t→∞
lifetime of the lightbulb is 100 hours.
9. The distance X (in cm) between a dart’s location on a dartboard and the bullseye (centre of the
dartboard) has probability density function
(
3
20x − x2
if 0 ≤ x ≤ 20
4000
f (x) =
0
otherwise
Calculate the standard deviation in X, σ(X).
First, the expected value is
Z
∞
E(X) =
Z
20
3
(20x2 − x3 )dx
4000
0
20
3
20 3 1 4 =
x − x 4000 3
4
0
xf (x)dx =
−∞
= 10
9
We also have that
2
Z
∞
E(X ) =
Z
20
3
(20x3 − x4 )dx
4000
0
20
1
3
5x4 − x5 =
4000
5
0
2
x f (x)dx =
−∞
= 120.
Therefore the variance is given by
2
Var(X) = E(X 2 ) = [E(X)] = 120 − 102 = 20,
√
and so the standard deviation is σ(X) = 20.
10