MATH 105 101
Midterm 2 Sample 2 Solutions
MATH 105 101 Midterm 2 Sample 2 Solutions
1. (20 marks)
(a) (5 marks) Find the derivative of the function:
Z x
πt
2
t sin
F (x) = x
dt,
6
3
at the point x = 3. Simplify the answer.
Solution: Applying product rule and the Fundamental Theorem of Calculus
Part I, we get:
Z x
Z x
πt
πt
dF
2 d
t sin
= 2x
dt + x
t sin
dt
dx
6
dx
6
3
3
Z x
πx πt
= 2x
t sin
dt + x2 x sin
.
6
6
3
Thus, at x = 3,
dF
|x=1 = 2(3)
dx
Z
3
t sin
3
πt
6
dt + 3
2
3 sin
π3
6
= 0 + 27 sin
π 2
= 27.
(b) (5 marks) Find a bound on the error if Simpson’s Rule is used to approximate
Z 5
ln(x) dx
1
with n = 4 subintervals.
Solution: We have a = 1, b = 5, n = 4, and f (x) = ln(x). So, ∆x =
To find an error bound, we first find the derivatives:
f 0 (x) = x−1 ,
f 00 (x) = −x−2 ,
b−a
n
= 1.
f 000 (x) = 2x−3 , f (4) (x) = −6x−4 .
So, |f (4) (x)| = 6x−4 . We want to find K such that for any x in [1, 5], then
|f (4) (x)| ≤ K. One way to find K is to find the maximum value of |f (4) (x)| on
[1, 5], we first find critical points on [1, 5]. Note that the derivative of 6x−4 is
−24x−5 which is never zero. So, the maximum can only occur at the endpoints.
Since |f (4) (1) = 6 and |f (4) (5)| = 6(5)−4 < 6, we may choose K = 6. Thus, a
bound for the error is:
24
K(b − a)(∆x)4
=
.
E4 ≤
180
180
Remark: We can also find K by making an observation that on [1, 5], we have
that 1 ≤ x, so 1 ≤ x4 . This means x−4 ≤ 1, and 6x−4 ≤ 6, so K = 6.
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MATH 105 101
Midterm 2 Sample 2 Solutions
(c) (5 marks) Find the definite integral
Z 5 √
3
− 25 − 9x2 dx.
−5
3
√
Solution: Let y = − 25 − 9x2 , then y 2 = 25 − 9x2 . In its proper form, the
equation:
y2
x2
+
=1
5 2
52
3
is an equation of an ellipse centered at (0, 0) with x-intercepts ± 35 , 0 and
√
y-intercepts (0, ±5). Moreover, y = − 25 − 9x2 corresponds to the lower
R 53 √
− 25 − 9x2 dx
semi-elliptic arc below the x-axis. The definite integral −5
3
√
represents the net area between the curve y = − 25 − 9x2 and the x-axis
and x = 35 , which corresponds to the lower half of the area of
between x = −5
3
, and half of it is 25π
.
the ellipse below the x-axis. The area of the ellipse is 25π
3
6
Since the area lies below the x-axis, its net area is negative, which means:
Z
5
3
−5
3
√
25π
− 25 − 9x2 dx = −
.
6
Remark: It is possible to use trigonometric substitution in this question, but it
will be much longer than necessary.
(d) (5 marks) Use summation identities to evaluate:
100
X
k(k + 2).
k=40
You may leave the answer unsimplified.
Solution: We first break the summation into two summations:
100
X
k(k + 2) =
k=40
100
X
2
(k + 2k) =
k=40
100
X
2
k +2
k=40
100
X
k.
k=40
To apply the summation identities, we need to transform the summation into
those with starting indices equal to 1, that is:
!
!
100
100
100
39
100
39
X
X
X
X
X
X
k2 + 2
k=
k2 −
k2 + 2
k−2
k .
k=40
k=40
k=1
k=1
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k=1
k=1
MATH 105 101
Midterm 2 Sample 2 Solutions
Apply the summation identities to each sum, we get:
!
!
100
100
39
100
39
X
X
X
X
X
k(k + 2) =
k2 −
k2 + 2
k−2
k
k=40
k=1
=
k=1
k=1
k=1
100(101)(201) 39(40)(79) 2(100)(101) 2(39)(40)
−
+
−
.
6
6
2
2
2. (10 marks) Evaluate the following definite integral:
Z 2
x3 + 6
dx.
2
1 x (x − 3)
Solution: First, note that the integrand is continuous on [1, 2], so we may use Fundamental Theorem of Calculus Part II. To do so, we want to find an antiderivative
3 +6
3 +6
3 +6
. Note that x2x(x−3)
= xx3 −3x
of x2x(x−3)
2 , and the numerator and the denominator
have the same degree. So, first we will perform polynomial long division, and get:
3x2 + 6
x3 + 6
=
1
+
.
x2 (x − 3)
x2 (x − 3)
We want to decompose
set:
−3x2 +6
x2 (x−3)
further using the method of partial fractions. We
A B
C
3x2 + 6
= + 2+
2
x (x − 3)
x x
x−3
2
A(x − 3x) + B(x − 3) + Cx2
=
x2 (x − 3)
2
(A + C)x + (B − 3A)x − 3B
=
.
x2 (x − 3)
⇒ 3x2 + 6 = (A + C)x2 + (B − 3A)x − 3B.
So, A + C = 3, B − 3A = 0, and 6 = −3B. Hence, B = −2, A = −2/3 and
C = 11/3. Thus,
Z Z
2
2
11
x3 + 6
dx =
1−
−
+
dx
x2 (x − 3)
3x x2 3(x − 3)
2
2 11
= x − ln |x| + +
ln |x − 3| + C.
3
x
3
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MATH 105 101
Then,
Z
1
2
Midterm 2 Sample 2 Solutions
x3 + 6
dx =
x2 (x − 3)
2
2 11
x − ln |x| + +
ln |x − 3| |21
3
x
3
2
11
2
11
= (2 − ln 2 + 1 +
ln 1) − (1 − ln 1 + 2 +
ln 2)
3
3
3
3
13
= − ln 2.
3
3. (10 marks) Find the solution in its explicit form of the initial value problem:
1
dy
cos2 (t) − = 0,
dt
y
y
π 4
= −2.
Solution: We have:
dy
1
dy
1
1
cos2 (t) − = 0 ⇔
cos2 (t) = ⇒ y dy =
dt.
dt
y
dt
y
cos2 (t)
Next, we want to integrate each side with respect to the respective variables. The
left hand side yields:
Z
y2
y dy =
+ C.
2
The right hand side yields:
Z
Z
1
dt = sec2 (t) dt = tan(t) + C.
cos2 (t)
So,
y2
2
= tan(t) + C. To find C, we use the initial condition y
π
4
= −2, and get:
(−2)2
π
= tan + C ⇒ C = 2 − 1 = 1.
2
4
So,
y2
2
= tan(t) + 1. To find the explicit form of y, we have:
y 2 = 2 tan(t) + 2 ⇒ y = ±
p
2 tan(t) + 2.
p
Since y π4 = −2 < 0, we get that y = − 2 tan(t) + 2 is the solution to the initial
value problem.
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MATH 105 101
Midterm 2 Sample 2 Solutions
4. (10 marks) Evaluate the indefinite integral:
Z
sec3 (x) dx.
Solution: Using integration by parts with u = sec(x) and dv = sec2 (x) dx, we get
du = sec(x) tan(x) dx and v = tan(x). Thus,
Z
Z
3
sec (x) dx = sec(x) tan(x) − sec(x) tan2 (x) dx.
R
To evaluate sec(x) tan2 (x) dx, observe that tan2 (x) = sec2 (x) − 1, so:
Z
Z
Z
Z
2
2
3
sec(x) tan (x) dx = sec(x)(sec (x) − 1) dx = sec (x) dx + sec(x) dx
Z
= sec3 (x) dx + ln | sec(x) + tan(x)|.
Replacing it into the first equation, we get:
Z
Z
3
sec (x) dx = sec(x) tan(x) − sec(x) tan2 (x) dx
Z
= sec(x) tan(x) − sec3 (x) dx − ln | sec(x) + tan(x)|.
Z
⇒ 2 sec3 (x) dx = sec(x) tan(x) − ln | sec(x) + tan(x)| + C
Z
1
1
⇒ sec3 (x) dx = sec(x) tan(x) − ln | sec(x) + tan(x)| + C.
2
2
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