APPM 1360
Exam 1 Solutions
Z
1.
2
Z
ln(t − 1) dt
(a) (10 pts)
(b) (10 pts)
Fall 2014
Z
x
e sin x dx
(c) (10 pts)
1
1
dx
x 4x2 + 16
√
(d) (12 pts) Verify the integration formula shown by integrating the left side of the equation and
showing that it equals the right side:
Z
1
1 x dx = ln + C, a, b > 0.
x(a + bx)
a
a + bx Solution:
(a) This is an improper integral since ln(t − 1) is undefined at t = 1.
Z 2
Z 2
ln(t − 1) dt = lim
ln(t − 1) dt
a→1+
1
a
Use integration by parts.
2
Z
= lim
a→1+
a
dt
ln(t − 1) |{z}
| {z }
u
dv
Then du = dt/(t − 1) and v = t.
!
t
= lim
dt
a→1+
a t−1
a
Z 2
1
dt
= lim 2 ln(1) − a ln(a − 1) −
1+
t−1
a→1+
a
!
2 Z
t ln(t − 1) −
2
2
= lim
a→1+
−a ln(a − 1) − t + ln|t − 1|
a
= lim (−a ln(a − 1) − [2 + ln(1) − a − ln(a − 1)])
a→1+
= lim (−a ln(a − 1) − 2 + a + ln(a − 1))
a→1+
= −2 + 1 + lim (−a ln(a − 1) + ln(a − 1))
a→1+
= −1 + lim (1 − a) ln(a − 1)
a→1+
This is an indeterminate product.
ln(a − 1)
(1 − a)−1
(a − 1)−1
LH
= −1 + lim
= −1 + lim (a − 1) = −1 + 0 = −1
a→1+ (1 − a)−2
a→1+
= −1 + lim
a→1+
(b)
using IBP with
u = ex
v = − cos x
Z
ex sin x dx = −ex cos x + cos x ex dx
du = e
Z
dv = sin x dx
x
using IBP again with
U = ex
dV = cos x dx
x
dU = e dx
Z
x
x
V = sin x
x
Z
e sin x dx = −e cos x + e sin x −
Z
ex sin x dx = ex sin x − ex cos x
Z
ex sin x dx =
2
sin x ex dx
ex
(sin x − cos x) + C
2
Alternate solution:
using IBP with
u = sin x
dv = ex dx
du = cos x dx v = ex
Z
Z
ex sin x dx = ex sin x − ex cos x dx
using IBP again with
U = cos x
Z
dV = ex dx
dU = − sin x dx V = ex
Z
x
x
x
x
e sin x dx = e sin x − e cos x + e sin x dx
Z
ex sin x dx = ex sin x − ex cos x
Z
ex sin x dx =
2
ex
(sin x − cos x) + C
2
(c)
Z
Z
1
1
√
dx =
dx
2
x 4x + 16
2x x2 + 4
Substitute x = 2 tan θ
√
=⇒ x2 = 4 tan2 θ
=⇒ x2 + 4 = 4 tan θ + 4
=⇒ x2 + 4 = 4(tan2 θ + 1)
Z
=⇒ x2 + 4 = 4 sec2 θ
Z
2 sec2 θ
1
√
dx =
dθ
4 tan θ(2 sec θ)
2x x2 + 4
Z
sec θ
=
dθ
4 tan θ
Z
1
cos θ
1
·
dθ
=
4
cos θ sin θ
Z
1
csc θ dθ
=
4
1
1
= − ln | csc θ + cot θ| + C or ln | csc θ − cot θ| + C
4
4
√
√
1 x2 + 4 2 1 x2 + 4 2 = − ln + + C or ln − +C
4 x
x
4 x
x
(d) Consider the integrand and use partial fraction decomposition.
1
A
B
= +
x(a + bx)
x
(a + bx)
Clear the fractions:
1 = A(a + bx) + Bx
If x = 0 then 1 = Aa =⇒ A = a1
−a
Similarly, if x = −a
b =⇒ 1 = B( b ) =⇒ B =
Therefore,
Z
1
−b
+
dx
ax a(a + bx)
Z
1
b
= ln |x| + −
dx
a
a(a + bx)
1
dx =
x(a + bx)
Z −b
a
The second integral can be solved using a U-substitution:
U = a + bx =⇒ dU = b dx
Z
Z
−b
1
dx = −
dU
a(a + bx)
aU
1
= − ln |U | + C
a
Z
1
1
1
dx = ln |x| − ln |a + bx| + C
x(a + bx)
a
a
1 x =
ln
+C
a a + bx 2. Determine if the following improper integrals are convergent or divergent. Cite any theorems or tests
that you use.
Z 1
Z +∞
Z +∞
cos x
ex
1 + e−2x
√ dx
(a) (9 pts)
(b) (9 pts)
dx
(c)
(9
pts)
dx
e2x + 1
x
x
0
0
1
Solution:
cos x
1
(a) For 0 < x ≤ 1, √ ≤ √ . Now
x
x
Z 1
Z 1
i1
h
√
1
√ dx = lim
I=
x−1/2 dx = lim 2x1/2 = lim 2 − 2 t = 2,
x
t
t→0+
t→0+ t
t→0+
0
Z 1
cos x
√ dx also is convergent .
so I is convergent. By the Comparison Test,
x
0
x
x
e
e
1
(b) For x ≥ 0, 2x
< 2x = x . Now
e +1
e
e
Z ∞
Z t
t
1
e−x dx = lim −e−x 0 = lim −e−t + 1 = 1,
I=
dx = lim
x
t→∞
t→∞
t→∞
e
0
0
Z +∞
ex
so I is convergent. By the Comparison Test,
dx also is convergent .
e2x + 1
0
Alternate solution:
Z
0
Let u =
ex , du
=
+∞
ex
dx = lim
t→∞
e2x + 1
Z
0
t
ex
dx
e2x + 1
ex dx.
Z
et
et
du
= lim tan−1 u 1
t→∞ 1
+ 1 t→∞
π π
π
= lim tan−1 et − tan−1 1 = − =
t→∞
2
4
4
= lim
The integral is convergent .
u2
1 + e−2x
1
(c) For x ≥ 1,
>
since e−2x > 0. The integral
xZ
x
+∞
1 + e−2x
dx also is divergent .
Comparison Test,
x
1
Z
∞
1
1
dx is divergent so by the
x
3. (15 pts) Find the area of the shaded region, shown at right,
bounded by
y = f (x) =
sin x
+ 1,
cos2 x
y = g(x) =
y
1
y ‡ f HxL
sin2 x
,
cos2 x
and the y-axis. The functions f and g intersect at y = 1/3.
1
3
y ‡ gHxL
0
x
Solution: First solve for the x-coordinate of the intersection point. Note that
sin x
sin2 x
+
1
=
sec
x
tan
x
+
1
and
g(x)
=
= tan2 x.
cos2 x
cos2 x
√
Then g(x) = tan2 x = 1/3 ⇒ tan x = −1/ 3 ⇒ x = −π/6. The functions f and g intersect at
x = −π/6. The area of the shaded region is
f (x) =
Z
0
Z
0
[f (x) − g(x)] dx =
−π/6
−π/6
0
Z
=
=
−π/6
Z 0
sec x tan x + 1 − tan2 x dx
sec x tan x + 1 − sec2 x − 1
dx
sec x tan x − sec2 x + 2 dx
−π/6
0
= sec x − tan x + 2x
−π/6
= (1 − 0 + 0) −
√2
3
+
√1
3
−
π
3
= 1−
√
3+
π
3
4. The following problems are not related.
Z 25
4
1
√
(a) (8 pts) Let g(x) =
+
. If the midpoint rule is used to approximate
g(x) dx,
3 x + 3 64x
1
estimate the minimum number of subintervals needed to produce an error less than 10−6 .
|EM | ≤
K(b − a)3
24n2
where |g 00 | ≤ K.
Solution: First find g 00 .
1
4
g(x) = (x + 3)−1/2 + x−1
3
64
2
1
0
−3/2
g (x) = − (x + 3)
− x−2
3
64
1
g 00 (x) = (x + 3)−5/2 + x−3
32
Since |g 00 | is a decreasing function, |g 00 | has a maximum value on [1, 25] at x = 1.
1
1
1
00
−3
−5/2
|g | = (x + 3)
+ x ≤ (4)−5/2 + (1) =
32
32
16
Let K = 1/16 and solve for n.
K(b − a)3
< 10−6
24n2
1
3
16 (25 − 1)
< 10−6
24n2
242
1
< 6
2
16n
10
2
24
n2 > 2 · 106
4
n > 6000
|EM | ≤
(b) (8 pts) Consider the graph of y = f (x), shown below. (No explanations are necessary for this
problem.)
Suppose the left, right, trapezoidal, and midpoint approximations are used to estimate
Z
a
f (x) dx with 3 subintervals.
0
i. Write the rules L3 , R3 , T3 , and M3 in order, from the smallest estimate to the largest.
ii. For each of the
Z four estimates, state whether it is an underestimate or an overestimate of the
a
f (x) dx.
true value of
0
Solution:
i. R3 , T3 , M3 , L3
ii. R3 and T3 are underestimates. M3 and L3 are overestimates.
y
0
L3
y
a
x
0
R3
y
a
x
0
T3
y
a
x
0
M3
a
x