Download Calculus II Exam 2 Solutions

Calculus II
Math 142 Winter 2005
Professor Ben Richert
Exam 2
Solutions
Problem 1. (10 pts.) A telephone line hanging between two poles has height y = 20 cosh(x/20)−15 meters above the ground.
What is the average height of the telephone line from x = −7 to x = 7?
Z b
1
f (x) dx. So the average height of the wire is
Solution. The equation for the average value of f (x) on [a, b] is
b−a a
Z 7
1
(20 cosh(x/20) − 15) dx. We use u-substitution: let u = x/20, so that du/dx = 1/20, and 20du = dx.
7 − (−7) −7
Furthermore, if x = −7, then u = −7/20, and if x = 7, then u = 7/20. So the average value is
Z 7/20
Z
1
20 7/20
(20 cosh u − 15)20 du =
(20 cosh u − 15)du
14 −7/20
14 −7/20
=
20
20
20
7/20
(20 sinh u − 15u|−7/20 =
(20 sinh(7/20) − 15(7/20)) − (20 sinh(−7/20) − 15(−7/20))
14
14
14
meters.
Z
∞
dx
, or show that it diverges.
(2x
+ 1)3
1
Z t
Z ∞
Z t
dx
dx
dx
Solution. We know that
= lim
. We can compute
using the u-substitution. Let
3
3
t→∞
(2x
+
1)
(2x
+
1)
(2x
+ 1)3
1
1
1
u = 2x + 1, so that du/dx = 2 and du/2 = dx. Then
x=t t
Z t
Z x=t
−1
−1
dx
du
−1 1
1
1
=
=
=
=
+
=
−
.
3
3
2
2
2
2
(2x
+
1)
2(u)
4u
4(2x
+
1)
4(2t
+
1)
4(3)
36
4(2t
+ 1)2
1
x=1
x=1
1
Problem 2. (10 pts.) Evaluate
So
Z
1
∞
dx
= lim
t→∞
(2x + 1)3
Z
t
1
dx
= lim
t→∞
(2x + 1)3
1
1
−
36 4(2t + 1)2
=
1
36
−1
because lim
= 0 (the denominator goes to infinity while the numerator is a constant). Thus the integral converges
t→∞ 4(2t + 1)2
1
to
.
36
Problem 3. (10 pts.) Compute the following limits.
(a – 5 pts.) lim x2 e−x
x→∞
f (x)
f 0 (x)
= lim 0
if both derivatives exist, g 0 (x) 6= 0, and lim f (x) =
x→∞ g(x)
x→∞ g (x)
x→∞
2
x
∞ = lim g(x) or lim f (x) = 0 = lim g(x). So note that x2 e−x = x , and that this quotient has the form
x→∞
x→∞
x→∞
e
∞
x2
2x
“ ” as x → ∞. Thus lim x = lim x . The hypotheses for L’Hospital’s rule still hold, so we repeat, yielding
x→∞ e
x→∞ e
∞
2x
2
lim
= lim x = 0 (because the denominator increases without bound as x increases while the numerator
x→∞ ex
x→∞ e
x2
is constant). We conclude that lim x = 0.
x→∞ e
Solution. L’Hospital’s rule states that lim
(b – 5 pts.) lim (x)1/x
x→∞
ln x
0
Solution. Let y = (x)1/x . Then ln y = ln(x)1/x = (1/x) ln x =
. This quotient has the form “ ” as x → 0+ ,
x
0
so we apply L’Hospital’s rule, and find that
lim ln y = lim
x→0+
x→0+
1
1
ln x
= lim x = lim
= 0.
+
+
x
x→0 1
x→0 x
Thus
lim y = lim+ eln y = elimx→0+ ln y = e0 = 1.
x→0+
x→0
Problem 4. (20 pts.) Integrate two of the following (indicate which two you want graded):
Z π/2
(a – 10 pts.)
x sin 2x dx
Z0
1
(b – 10 pts.)
dx
2+x−6
x
Z
p
(c – 10 pts.)
x3 x2 − 1 dx
Solution. We can compute by first doing a u-substitution, and then using integration by parts. Let u = 2x, so that du/dx = 2,
or du/2 = dx and x = u/2. Furthermore, when x = 0, then u = 0 and when x = π/2, then u = π. So
Z π/2
Z
Z π
du
1 π
u
sin u
=
x sin 2x dx =
u sin u du.
2
4 0
0
0 2
We now do integration by parts. Use the table:
f = u g = − cos u
f 0 = 1 g 0 = sin u
Z
Z
Z
Z π
1 π
1
and the formula
f g 0 dx = f g − f 0 g dx. So we get that
u sin u du =
u(− cos u)|π0 −
− cos u du =
4 0
4
0
Z π
1
1
π
1
π
π(− cos π) − (0) +
cos u du = (π(−(−1)) + sin u|0 ) = (π + sin π − sin 0) = .
4
4
4
4
0
Solution. We can compute the second integral by expanding the fraction. We know that
1
1
A
B
=
=
+
x2 + x − 6
(x − 2)(x + 3)
x−2 x+3
for some numbers A, B. Multiplying both sides of this equation by (x − 2)(x + 3) yields the equality
1 = A(x + 3) + B(x − 2).
This means that
1 = (A + B)x + (3A − 2B)
and thus that A + B = 0 (so that A = −B) and 3A − 2B = 1. Substituting A = −B into the second equation, we find that
1
1
3(−B) − 2B = 1, or that −5B = 1, so that B = − . It follows that A = . Thus
5
5
Z
Z 1
Z
Z
1 −5
1
1
1
1
1
5
dx
=
+
dx
=
dx
−
dx.
2
x +x−6
x−2 x+3
5
x−2
5
x+3
We now use a u-substitution on the first integral and a w-substitution on the second. Let u = x − 2 whence du = dx and
w = x + 3 whence dw = dx. So
Z
Z
Z
Z
1
1
1
1
1
1
1
1
1
1
1
1
dx −
dx =
du −
dw = ln |u| − ln |w| + C = ln |x − 2| − ln |x + 3| + C.
5
x−2
5
x+3
5
u
5
w
5
5
5
5
Solution. For the third integral, we use a trig substitution. Let x = sec θ, so that dx = sec θ tan θ dθ. Thus
Z
Z
Z
p
p
p
x3 x2 − 1 dx = sec3 θ sec2 θ − 1 sec θ tan θ dθ = sec4 θ tan2 θ tan θ dθ
Z
Z
Z
Z
= sec4 θ tan θ tan θ dθ = sec4 θ tan2 θ dθ = sec2 sec2 θ tan2 θ dθ = sec2 θ(tan2 θ + 1) tan2 θ dθ.
So let u = tan θ. Then du = sec2 θ dθ, and we can rewrite
Z
sec2 θ(tan2 θ + 1) tan2 θ dθ
as
Z
(u2 + 1)u2 du =
Z
(u4 + u2 ) du = u5 /5 + u3 /3 + C = (tan θ)5 /5 + (tan θ)3 /3 + C.
Now because x = sec θ, we know that sec−1 x = θ, so √
tan sec−1 x = tan θ. As secant is hypotenuse over opposite and tangent
p
p
x2 − 1
, and the final answer is ( x2 − 1)5 /5 + ( x2 − 1)3 /3 + C. is opposite over adjacent, we can see that tan sec x =
1
Z 6
Problem 5. (16 pts.) Consider the definite integral
x3 + 1 dx.
0
(a – 8 pts.) Estimate this integral using the trapezoidal rule with n = 3.
Z 6
Solution. The formula for the trapezoidal rule is
x3 + 1 dx ≈ ∆x/2(f (x0 ) + 2f (x1 ) + · · · + 2f (xn−1 ) + f (xn )).
0
In our case, we take n = 3, so ∆x = 6/3 = 2, x0 = 0, x1 = 2, x2 = 4, and x3 = 6. So our estimate is
2/2((03 + 1) + 2(23 + 1) + 2(43 + 1) + (63 + 1)) = (1 + 2(9) + 2(65) + (217)) = 366.
(b – 8 pts.) Give a bound for the error in your answer (use an the error formula, not the Fundamental Theorem of Calculus).
K(b − a)3
where K is such that |f 00 (x)| ≤ K for all a ≤ x ≤ b. In our situation, we note
12n2
that f 0 (x) = 3x2 and f 00 (x) = 6x. This is clearly an increasing function (its derivative is always positive), so it takes
it absolute max and absolute min at the endpoints. It’s absolute max on [0, 6] is f 00 (6) = 36, while its absolute min
is f 00 (0) = 0. So if we let K = 36, it will be true that |f 00 (x)| ≤ K for all 0 ≤ x ≤ 6. Thus we conclude that the error
K(b − a)3
36(6)3
is bounded by
=
= 72. (As a side note, we can compute the actual error using the fundamental
2
12n
12(32 )
Z 6
theorem: we have that ET =
(x3 + 1) dx − 366 = (x4 /4 + x|60 − 366 = 64 /4 + 6 − 366 = −36; so we see that the
Solution. Recall that |ET | ≤
0
absolute value of our error is bounded by 72 as we calculated).
Problem 6. (14 pts.) Each of the integrals on the right corresponds to exactly one integral on the left (after a u-substitution,
a trig-substitution, or some other manipulation). Match the integrals.
Z
Z
sin 3x cos 2x dx
E
Z
Z
F
sin x cos(cos x) dx
Z
A
(a)
(b)
tan5 x sec3 x dx
Z
2x cot x2 sin x2 dx
Z
C
x2
x
dx
+ 6x + 8
x2
x
dx
+ 3x + 2
Z
G
Z
B
√
√
u
du
u−1
Z
(c)
D
u2 (u2 − 1)2 du
(
2
−1
+
) dx
x+2 x+4
Z
(d)
cos u du
1
(e)
2
Z
(sin x + sin 5x) dx
Z
(f ) −
1 + ex dx
cos u du
Z
(g)
(
−1
2
+
) dx
x+1 x+2