MATH 105 101
Midterm 2 Sample 3 Solutions
MATH 105 101 Midterm 2 Sample 3 Solutions
1. (20 marks)
dF
dx
of the function:
Z x
sin(t) dt.
F (x) =
(a) (5 marks) Find the derivative
cos(x)
Solution: We have that:
Z
Z x
sin(t) dt = −
F (x) =
cos(x)
cos(x)
Z
sin(t) dt +
0
x
sin(t) dt.
0
Applying the Fundamental Theorem of Calculus Part I, and chain rule, we get:
dF
= − sin(cos(x))(− sin(x)) + sin(x) = sin(cos(x)) sin(x) + sin(x).
dx
(b) (5 marks) Use the Trapezoidal Rule to approximate
Z 6
(x − 2)2 dx
0
with n = 3 equal subintervals. Simplify your answer.
Solution: We have a = 0, b = 6, n = 3, and f (x) = (x − 2)2 . So, ∆x =
2. There are 4 grid-points using the formula xk = a + k∆x:
x0 = 0,
x1 = 2,
x2 = 4,
x3 = 6.
Using Trapezoidal Rule, we get:
∆x
(f (x0 ) + 2f (x1 ) + 2f (x2 ) + f (x3 ))
2
2
= ((0 − 2)2 + 2(2 − 2)2 + 2(4 − 2)2 + (6 − 2)2 )
2
= 4 + 0 + 8 + 16 = 28.
T3 =
(c) (5 marks) Evaluate the indefinite integral:
Z
cos5 (x) dx.
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b−a
n
=
MATH 105 101
Midterm 2 Sample 3 Solutions
Solution: Using the identity cos2 (x) = 1 − sin2 (x), we get:
Z
Z
Z
5
4
cos (x) dx = cos (x) cos(x) dx = (1 − sin2 (x))2 cos(x) dx.
Using substitution with u = sin(x) and du = cos(x) dx, we get:
Z
Z
Z
2u3 u5
2
2
2 2
+
+ C.
(1 − sin (x)) cos(x) dx = (1 − u ) du = (1 − 2u2 + u4 ) du = u −
3
5
Hence,
Z
cos5 (x) dx = sin(x) −
2 sin3 (x) sin5 (x)
+
+ C.
3
5
(d) (5 marks) Compute the Midpoint Riemann sum for the function f (x) = x2 on the
interval [−5, 5] using n = 5 equal subintervals. Simplify your answer.
Solution: We have a = −5, b = 5, and n = 5. So, ∆x =
Midpoint Riemann sum, we have:
b−a
n
= 2. For
x∗k = a + (k − 1/2)∆x = −5 + (k − 1/2)2 = −6 + 2k.
Since n = 5, the five midpoints x∗k are:
x∗1 = −4,
x∗2 = −2,
x∗3 = 0,
x∗4 = 2,
x∗5 = 4.
Thus, The Midpoint Riemann sum for f (x) = x2 on [−5, 5] using n = 5 equal
subintervals is:
f (x∗1 )∆x + f (x∗2 )∆x + f (x∗3 )∆x + f (x∗4 )∆x + f (x∗5 )∆x
= 2(−4)2 + 2(−2)2 + 2(02 ) + 2(22 ) + 2(42 )
= 32 + 8 + 0 + 8 + 32 = 80.
2. (10 marks) Evaluate the following indefinite integral:
Z √ 2
x − 2x − 8
dx.
x−1
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MATH 105 101
Midterm 2 Sample 3 Solutions
Solution: Completing the square, we get x2 − 2x − 8 = (x − 1)2 − 32 . So, let
u = x − 1, and du = dx, we get:
Z √ 2
Z √ 2
x − 2x − 8
u − 32
dx =
du.
x−1
u
We
√ use trigonometric substitution with u = 3 sec(θ), du = 3 sec(θ) tan(θ) dθ. Then,
u2 − 32 = 3 tan(θ), and:
Z
Z √ 2
u − 32
3 tan(θ)
du =
(3 sec(θ) tan(θ) dθ)
u
3 sec(θ)
Z
Z
2
= 3 tan (θ) dθ = 3 (sec2 (θ) − 1) dθ
Z
Z
2
= 3 sec (θ) dθ − 3 dθ
= 3 tan(θ) − 3θ + C.
Since u = 3 sec(θ), we get
Z √
3
u
= cos(θ), and θ = arccos
√
x2 − 2x − 8
dx = u2 − 9 − 3 arccos
x−1
3
u
. So,
p
3
3
+ C = (x − 1)2 − 9 − 3 arccos
+ C.
u
x−1
3. (10 marks) Find the solution of the initial value problem:
dy −y ln(t)
e −
= 0,
dt
t
You may leave the answer in its implicit form.
y(1) = 0.
Solution: We have:
dy −y ln(t)
dy
ln(t)
ln(t)
e −
= 0 ⇔ e−y =
⇒ e−y dy =
dt.
dt
t
dt
t
cos2 (t)
Next, we want to integrate each side with respect to the respective variables. The
left hand side yields:
Z
e−y dy = −e−y + C.
For the integral
R
ln(t)
t
dt, we use substitution with x = ln(t) and dx = 1t dt. Then,
Z
Z
x2
ln(t)
ln2 (t)
dt = x dx =
+C =
+ C.
t
2
2
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MATH 105 101
So, −e−y =
Midterm 2 Sample 3 Solutions
ln2 (t)
2
+ C. To find C, we use the initial condition y(1) = 0, and get:
−e0 =
ln2 (1)
+ C ⇒ C = −1.
2
So, the solution to the initial value problem in its implicit form is:
−e−y =
ln2 (t)
− 1.
2
4. (10 marks) Evaluate the definite integral:
Z 8 3
x − 6x2 − 2x − 11
dx.
x2 − 6x − 7
0
Solution: First, observe that the integrand is undefined at x = −1 and x = 7, so
this is an improper integral, and:
Z 8 3
Z a 3
Z 8 3
x − 6x2 − 2x − 11
x − 6x2 − 2x − 11
x − 6x2 − 2x − 11
dx = lim−
dx + lim+
dx,
a→7
b→7
x2 − 6x − 7
x2 − 6x − 7
x2 − 6x − 7
0
0
b
if both limits exist. First, we want to find an antiderivative of the integrand.
Observe that the degree of the denominator is less than that of the numerator, so
first we need to do polynomial long division:
5x − 11
x3 − 6x2 − 2x − 11
=x+ 2
.
2
x − 6x − 7
x − 6x − 7
We will use the method of partial fractions to decompose the remaining fraction
further since x2 − 6x − 7 = (x + 1)(x − 7). We get:
5x − 11
A
B
A(x − 7) + B(x + 1)
=
+
=
(x + 1)(x − 7)
x+1 x−7
(x + 1)(x − 7)
(A + B)x + (B − 7A)
=
x2 − 6x − 7
⇒ 5x − 11 = (A + B)x + (B − 7A).
So, A + B = 5 and B − 7A = −11. Since B = 5 − A, we get −11 = B − 7A =
5 − A − 7A = 5 − 8A, which yields A = 2 and thus B = 3. Hence,
Z 3
Z x − 6x2 − 2x − 11
2
3
x2
dx
=
x
+
+
dx
=
+ 2 ln |x + 1| + 3 ln |x − 7| + C.
x2 − 6x − 7
x+1 x−7
2
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MATH 105 101
Midterm 2 Sample 3 Solutions
Thus,
Z
lim−
a→7
0
a
x3 − 6x2 − 2x − 11
x2
dx
=
lim
+ 2 ln |x + 1| + 3 ln |x − 7| |a0
2
−
a→7
x − 6x − 7
2
2
a
+ 2 ln |a + 1| + 3 ln |a − 7| − 3 ln 7
= lim−
a→7
2
49
=
+ 2 ln 8 − 3 ln 7 + lim− 3 ln |a − 7|.
a→7
2
As a → 7− , then |a − 7| → 0+ , so lim− 3 ln |a − 7| = −∞. Since one of the limit
a→7
does not exist, the improper integral:
Z 8 3
x − 6x2 − 2x − 11
dx
x2 − 6x − 7
0
diverges.
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