Download Solutions to Midterm I Problem 1. Find the limits: (a) lim (cos x) (b

Math 112 – Intermediate Calculus II
Fall 2008
Solutions to Midterm I
Problem 1. Find the limits:
x
(a) lim (cos x) x−sin x .
x→0
(b) lim
x→∞
cosh x + f (x)
, assuming that f (x) grows slower than ex when x → ∞.
sinh x − f (x)
solution: (a) Denote the limit in question by A. Then
ln cos x − x tan x h 0 i
− tan x − tan x − x sec2 x
x ln cos x h 0 i
= lim
= lim
=
=
= −3
x→0
x→0
x→0 x − sin x
0
1 − cos x
0
sin x
ln A = lim
(cancelling sin x and using limx→0 (x/ sin x) = 1). Hence, A = e−3 .
(b) Spell out and divide by ex :
1 + e−2x
f (x)
+ x
2
e
lim
= 1
x→∞ 1 − e−2x
f (x)
− x
2
e
Z
Problem 2. Evaluate
x3
+
f (x)
= 0).
x→∞ ex
(using lim e−2x = 0 and lim
x→∞
x
dx.
+ 8x + 8
4x2
solution: Note that x = −2 is a root of the denominator; then, dividing by (x + 2), we get x3 + 4x2 + 8x + 8 =
(x + 2)(x2 + 2x + 4). The partial fraction expansion is
A
Bx + C
x
=
+
,
x3 + 4x2 + 8x + 8
x + 2 x2 + 2x + 4
or x = A(x2 + 2x + 4) + (Bx + C)(x + 2).
Letting x = −2, we get A = − 21 . Letting x = 0, we get 0 = 4A + 2C, or C = 1. Finally, equating the coefficients
of x2 (the highest power of x), we get 0 = A + B, or B = 21 . In addition, notice that x2 + 2x + 4 = (x + 1)2 + 3
(complete the square) and 12 x + 1 = 12 (x + 1) + 12 . Hence, the integral in question is1
1
−
2
Z
dx
1
+
x+2 2
Z
x+1
1
dx +
2
x + 2x + 4
2
Z
√
1
3
dx
1
x+1
2
.
= − ln|x + 2| + ln(x + 2x + 4) +
tan−1 √
2
(x + 1) + 3
2
4
6
3
Problem
Z 3. Evaluate:
(a) sinh ln(x + 1) − ln(x − 1) dx.
Z
(b)
x sinh−1 (x2 ) dx.
solution: (a) First, simplify:
x+1
x+1
x+1
1
1
1x + 1 x − 1
1
1
sinh ln(x + 1) − ln(x − 1) = sinh ln
= eln x−1 − e− ln x−1 =
−
=
+
.
x−1
2
2
2 x−1 x+1
x−1 x+1
Then the integral is ln|x − 1| + ln|x + 1| = ln|x2 − 1| .
(b) First, substitute t = x2 , dt = 2x dx. Then the integral in question is
1
2
Z
"
−1
sinh
t dt =
u = sinh−1 t
v=t
1I
#
Z
dt
1
1
t dt
1
1p 4
−1
√
= x2 sinh−1(x2 ) −
x +1 .
t2 + 1 = 2 t sinh t − 2
2
2
2
t +1
dv = dt
du = √
am skipping the + C part to save space
Math 112 – Intermediate Calculus II
Solutions to Midterm I
Z tan x
π
d
dt
Problem 4. (a) Find g
, where g(x) =
dt.
2
6
dx − tan x (t + 1)2
1 (b) Express sech−1 √
in terms of the natural logarithm.
2
solution: (a) Use the fundamental theorem of calculus and the chain rule:
g(x) =
1
1
(tan x)0 −
(− tan x)0 = 2 cos2 x.
(tan2 x + 1)2
(tan2 x + 1)2
√ 3 2
3
=
=2
.
6
2
2
1 √
√
√
(b) Let x = sech−1 √ . Then cosh x = 2, i.e., ex + e−x = 2 2. Let ex = y, so that y 2 − 2 2y + 1 = 0, or
2
√
√
√
y = 2±1. Since we want x ≥ 0 (the definition of sech−1 ), we need y ≥ 1, i.e., y = 2+1 and x = ln( 2 + 1) .
Then g
π
d
(sin x)ln x .
Problem 5. (a) Find
dx
Z 2 Z x
2
et dt dx.
(b) Evaluate
0
2
solution: (a) Use logarithmic differentiation:
ln(sin x)ln x
0
0
ln(sin x)
+ ln x cot x,
= ln x ln(sin x) =
x
and
(sin x)ln x
0
= (sin x)ln x
ln(sin x)
+ ln x cot x .
x
(b) Integrate by parts (keeping in mind the fundamental theorem of calculus)
Z 2 Z
0
2
x
Rx 2
2
u = 2 et dt
et dt dx =
v=x
Z x
2 Z 2
2
h 1 2 i2
2
2
1 − e4
du = ex dx
=
.
= x
et dt −
xex dx = 0 − ex
dv = dx
2
2
0
2
0
0
R x 2
R2
R0
(In the last integral, we use the substitution x2 = t. The exintegral term is x 2 0 = 2 2 −0 2 = 0.)