Solutions to Sample Questions for Exam #1
1 Write down a Riemann Sum with n rectangles for the area above the x-axis and under
the curve y = sin(x) for 0 ≤ x ≤ π2 . Use right endpoints and Σ notation for your answer.
The interval 0 ≤ x ≤
π
2
has length π2 , so each subinterval will have length
π
∆x =
.
2n
The right endpoint of the ith subinterval is
xi = i∆x =
πi
.
2n
Thus the Riemann sum is
n
X
f (xi )∆x =
i=1
n
X
sin
i=1
πi
2n
π
.
2n
2 Estimate the area under the graph of y = ln x and above the x − axis for 1 ≤ x ≤ 2 using
4 subintervals and (a) right endpoints as sample points, and (b) left endpoints as sample
points. Include 3 decimal places in your answers.
(a) The width of the entire interval is 1, so the width of each subinterval is ∆x = 14 . The
right endpoints of the subintervals are x1 = 54 , x2 = 32 , x3 = 47 and x4 = 2. Therefore, the
areas of the approximating rectangles are
1 5
1 3
1 7
ln , A2 = ln , A3 = ln ,
4 4
4 2
4 4
Therefore the total area is approximately
A1 =
A1 + A2 + A3 + A4 =
and
A4 =
1
ln 2.
4
1 5 1 3 1 7 1
ln + ln + ln + ln 2 ≈ 0.470
4 4 4 2 4 4 4
(b) The width of the entire interval is 1, so the width of each subinterval is ∆x = 14 . The
left endpoints of the subintervals are x1 = 1, x2 = 54 , x3 = 23 and x4 = 74 . Therefore, the
areas of the approximating rectangles are
1
1 5
ln 1, A2 = ln ,
4
4 4
Therefore the total area is approximately
A1 =
A1 + A2 + A3 + A4 = 0 +
A3 =
1 3
ln ,
4 2
and A4 =
1 7
ln .
4 4
1 5 1 3 1 7
ln + ln + ln + ≈ 0.297
4 4 4 2 4 4
3 Calculate
R9
4
3x−2
√
x
dx. Show all your work.
Z
9
4
3x − 2
√
dx =
x
Z
=
Z
9
4
3x
2
√ − √ dx
x
x
9
1
1
3x 2 − 2x− 2 dx
4
3
1 9
2
2
= 2x − 4x 4
3
1
3
1
= 2 · 92 − 4 · 92 − 2 · 42 − 4 · 42
= (54 − 12) − (16 − 8)
= 34.
4 Calculate
R
x3 cos (x4 ) dx. Show all your work.
Z
3
4
x cos x
dx =
=
=
=
5 Calculate
R
Z
du
cos u
4
Z
1
cos u du
4
1
sin u + C
4
1
sin x4 + C.
4
du
3
u=x ,
= x dx
4
4
cot θ dθ using substitution. Show all your work.
Z
cos θ
dθ
sin θ
Z
1
=
du
(u = sin θ, du = cos θ dθ)
u
= ln |u| + C
= ln | sin θ| + C.
cot θ dθ =
Z
6 Calculate
Z
R4√
1
√
t ln t dt. Show all your work.
Z
1
t 2 ln t dt
Z 2 3
2 3 1
= ln t
t2 −
t2
dt
3
3
t
Z
1
2 3
2
2
= t ln t −
t 2 dt
3
3
2 3
4 3
= t 2 ln t − t 2
3
9
t ln t dt =
1
u = ln t dv = t 2 dt
3
du = 1t dt v = 32 t 2
Therefore
Z
4
1
7 Calculate
Z
R
√
4
2 3
4 3 t ln t dt = t 2 ln t − t 2 3
9 1
2 3
4 3
2 3
4 3
2
2
2
2
· 4 ln 4 − · 4
· 1 ln 1 − · 1
=
−
3
9
3
9
16 ln 4 28
−
=
3
9
x2 ex dx. Show all your work.
u = x2
dv = ex dx
x e dx = x e − 2xe dx
du = 2x dx
v = ex
Z
u = 2x
dv = ex dx
2 x
x
x
= x e − 2xe − 2e dx
du = 2 dx
v = ex
2 x
2 x
Z
x
= x2 ex − (2xex − 2ex ) + C
= x2 ex − 2xex + 2ex + C
8 Use partial fractions to evaluate
R
1
x2 +4x+3
dx. Show all your work.
The denominator x2 + 4x + 3 can be factored as (x + 3)(x + 1), so we want to decompose
the fraction as follows:
A
B
1
=
+
(x + 3)(x + 1)
x+3 x+1
Multiply both sides by the denominator of the left to get
1 = A(x + 1) + B(x + 3).
If we plug in x = −1 we get the equation
1 = 2B,
so B = 12 . If we plug in x = −3 we get the equation
1 = −2A,
so A = − 12 . That is to say,
1
− 12
1
=
+ 2 .
(x + 3)(x + 1)
x+3 x+1
Consequently,
Z
Z
1
1
dx
=
dx
x2 + 4x + 3
(x + 3)(x + 1)
Z
1
− 12
=
+ 2 dx
x+3 x+1
Z
Z
1
1
1
1
=−
dx +
dx
2
x+3
2
x+1
Z
Z
1
1
1
1
=−
ds +
dt (s = x + 3, ds = dx and t = x + 1, dt = dx)
2
s
2
t
1
1
= − ln |s| + ln |t| + C
2
2
1
1
= − ln |x + 3| + ln |x + 1| + C.
2
2
9 First use substitution, then integration-by-parts, to calculate the integral
Z
√
cos x dx =
Z
(cos s)2s ds
(s =
Z
= 2s sin s − 2 sin s ds
√
R
√
cos x dx.
1
x, ds = √ dx, or 2s ds = dx)
2 x
u = 2s
dv = cos s ds
du = 2 ds
v = sin s
= 2s sin s + 2 cos s + C
√
√
√
= 2 x sin x + 2 cos x + C.
10 Calculate the area under the curve y = x2 and above the x-axis, for 0 ≤ x ≤ 3, by
calculating a limit of Riemann Sums. You may use the following formula:
N
X
j=1
j2 =
N(N + 1)(2N + 1)
.
6
With N rectangles, the length of each subinterval is ∆x =
. Therefore the area is equal to
sample points wil be xj = 3j
N
3
.
N
Using right endpoints, the
n
X
2
n X
3j
3
lim
f (xj )∆x = lim
N →∞
N →∞
N
N
j=1
j=1
= lim
N →∞
n
X
27j 2
j=1
N3
n
27 X 2
= lim 3
j
N →∞ N
j=1
27 N(N + 1)(2N + 1)
N →∞ N 3
6
18N 2 + 27N + 9
= lim
N →∞
2N 2
= 9.
= lim
11 A function f (x) is defined on the interval (0, ∞) by the formula
f (x) =
Z
x2
0
sin(t)
dt.
t
Find all the critical points of this function.
According to the Fundamental Thoerem of Calculus,
Z x2
sin(t)
sin(x2 ) d
d
2
dt =
x
dx 0
t
x2 dx
sin(x2 )
=
(2x)
x2
2 sin(x2 )
=
.
x
So the critical points occur where this is zero:
2 sin(x2 )
=0
x
=⇒
where k is any positive integer.
sin(x2 ) = 0
=⇒
x2 = πk
=⇒
x=
√
πk ,