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Solution: APPM 1360
Exam 1
Fall 2015
1. The following parts are not related:
Z π/4
(a)(10 pts) Evaluate the integral
x sec2 (x) dx
0
x2 − x + 2
dx
(x − 1)(x2 + 1)
(c)(5 pts) Suppose f (x) is a differentiable function defined on the interval 0 < a ≤ x ≤ b and suppose af (a) = bf (b),
Z b
Z b
prove that
f (x)dx = −
xf 0 (x) dx.
Z
(b)(10 pts) Evaluate the integral using the method of partial fractions:
a
a
Solution: (a)(10 pts) Using integration by parts with u = x ⇒ du = dx and dv = sec2 (x)dx ⇒ v = tan(x), we
have
Z
0
π/4
π/4 Z
x sec (x) dx = x tan(x)
−
π/4
2
0
tan(x) dx
0
and note that
π/4
π/4
π/4
Z π/4
√
π
= ln( 2)
= ln | sec(x)|
tan(x) dx = − ln | cos(x)|
= and
x tan(x)
4
0
0
0
0
so,
Z
√
π
− ln( 2)
4
π/4
x sec2 (x) dx =
0
(b)(10 pts) Using partial fractions, we have
A
Bx + C
x2 − x + 2
=
+ 2
=⇒ x2 − x + 2 = A(x2 + 1) + (Bx + C)(x − 1) ⇒ A = 1, B = 0, C = −1
2
(x − 1)(x + 1)
x−1
x +1
thus,
Z
x2 − x + 2
dx =
(x − 1)(x2 + 1)
Z
1
dx −
x−1
Z
x2
1
dx = ln |x − 1| − tan−1 (x) + C
+1
(c)(5 pts) Using integration by parts with u = f (x) ⇒ du = f 0 (x) and dv = dx ⇒ v = x, we have
Z
a
b
b Z
f (x)dx = xf (x) −
a
b
Z
0
xf (x) dx = bf (b) − af (a) −
a
b
f (x)dx
a
Z
and note that af (a) = bf (b) implies bf (b) − af (a) = 0, so
b
Z
f (x)dx = −
a
b
f (x)dx.
a
2. Determine if the integral converges or diverges, find the value of the integral if it converges, show all work:
Z 4
Z ∞
ln(x)
dx
√ dx
dx
(a)(6 pts)
(b)(6 pts)
(x + 3)4
x
0
1
Determine if the integral converges or diverges, justify your answer:
Z ∞
Z 1
sin4 (x) + e−x
sec2 (x)
√ dx
(c)(6 pts)
dx
(d)(6
pts)
2
x
x x
1
0
(Fun fact! You may or may not be interested to know that 1 < e < 3.)
Solution:
(a)(6 pts) We first write the improper integral as a limit
Z 4
Z 4
ln(x)
ln(x)
√ dx = lim
√ dx
+
x
x
t→0
0
t
√
now using integration by parts with u = ln(x) ⇒ du = dx/x and dv = x−1/2 dx ⇒ v = 2 x, we have
Z
Z
√
√
√
2
ln(x)
√ = 2 x ln(x) −
√ dx = 2 x ln(x) − 4 x + C
x
x
and so
4
Z
lim+
t→0
t
√
√ L0 H
√
√ 4
ln(x)
√ dx = lim 2 x ln(x) − 4 x = lim 4 ln(4) − 8 − 2 t ln(t) + 4 t = 4 ln(4) − 8
+
+
x
t→0
t→0
t
where in the last equality we used L’Hospital’s Rule, i.e. note that
√
2/t
2 ln(t) L0 H
0·∞
lim+ 2 t ln(t) = lim+ −1/2 = lim+ −3/2 = lim+ −4t1/2 = 0
t→0
t→0 −t
t→0 t
/2 t→0
Thus, the integral converges to 4 ln(4) − 8 .
(b)(6 pts) Writing the integral as a limit and using u-substitution yields
∞
Z
1
dx
dx = lim
b→∞
(x + 3)4
b
dx
1
1
1
1
1
dx
=
lim
−
=
=
=
lim
−
+
.
b→∞
(x + 3)4
3(x + 3)3 1 b→∞
3(b + 3)3 3(4)3
3(4)3
192
|
{z
}
b
Z
1
→0
Thus, the integral converges to 1/192 .
(c)(6 pts) We classify the improper integral by comparison. Note that x ≥ 1 implies ex ≥ e and so for all x ≥ 1
we have that e−x ≤ 1/e < 1, also recall e−x > 0, and so for x ≥ 1 we have
0 ≤ sin4 (x) ≤ 1
Z
0 < sin4 (x) + e−x ≤ 1 + 1/e < 2
=⇒
=⇒
0<
1
Z
and note that
1
∞
2
dx = 2
x2
∞
Z
1
1
dx is convergent since
x2
Z
1
∞
∞
sin4 (x) + e−x
dx <
x2
Z
1
∞
2
dx
x2
1
dx is a convergent p-integral with p = 2 > 1.
x2
∞
sin4 (x) + e−x
dx converges by Direct Comparison Test.
So,
x2
1
(d)(6 pts) We classify the improper integral by comparison. Note that for 0 ≤ x ≤ 1 we have cos(x) ≤ 1 and so
Z
cos2 (x) ≤ 1 and thus sec2 (x) ≥ 1 for 0 ≤ x ≤ 1 and so
Z
0
Z
So
0
1
1
sec2 (x)
√ dx ≥
x x
Z
0
1
1
x3/2
Z
1
dx > 0, and note
0
1
x3/2
dx = lim+ −2x
t→0
1
2
lim+ −2 + √ = +∞
= t→0
t
t
−1/2 sec2 (x)
√ dx diverges by Direct Comparison Test.
x x
3. The following problems are not related:
(a)(8 pts) For the given integrals, write down
substitution required to solve the
Z p the appropriate trigonometric
Z
dx
integral, do not solve the integrals: (i)
4x2 − 25 dx
(ii)
x2 + 4x + 13
Z
(b)(8 pts) Find the antiderivative:
sin2 (θ) dθ (Give your final answer in terms of the reference angle θ.)
Z 1 √
1 − x2
(c)(9 pts) Solve the following integral:
dx
x2
1/2
Solution:
p
√
(a)(i)(4 pts) Here we have a difference of squares, 4x2 − 25 = (2x)2 − 52 , so we could use either one of the
5
5
following substitutions: x = sec(θ) or x = csc(θ) .
2
2
(a)(ii)(4 pts) Here, after completing the square in the denominator, we have x2 + 4x + 13 = (x + 2)2 + 32 , i.e. a
sum of squares, so we could use either x = 3 tan(θ) − 2 or x = 3 cot(θ) − 2 .
1 − cos(2θ)
we have
2
Z
Z
1
sin(2θ)
1
2
(1 − cos(2θ)) dθ =
θ−
+C
sin (θ) dθ =
2
2
2
1
2 sin(θ) cos(θ)
θ − sin(θ) cos(θ)
=
θ−
+C =
+C
2
2
2
(b)(8 pts) Using the identity sin2 (θ) =
note that in the work above we used the identity sin(2θ) = 2 sin(θ) cos(θ).
(c)(9 pts) Using the trigonometric substitution x = sin(θ) ⇒ dx = cos(θ)dθ and so we have,
Z π/2
Z 1 √
Z π/2
cos2 (θ)
1 − x2
dx =
dθ =
cot2 (θ) dθ
2
x2
π/6 sin (θ)
1/2
π/6
Z π/2
=
(csc2 (θ) − 1) dθ
π/6
=
π/2
√
√
π
= [−0 − π/2] − [− 3 − π/6] = 3 −
− cot(θ) − θ
3
π/6
4. The following problems are not related:
(a)(8 pts) Find the area of the region enclosed by f (x) = x2 and g(x) = x2 ln(x), and the lines x = 1 and x = e.
Z 1
(b)(8 pts) Suppose we approximate
sin(πx) dx ≈ T4 , find a bound for |ET |, the absolute error of this approxi0
mation without evaluating the integral or finding the approximation.
Z
2
(c)(10 pts) The area between the curves y = x + 1 and y = x − x + 1 is given by the integral
2
(2x − x2 )dx.
0
Suppose we approximate this area using the Midpoint Rule with n partitions. What is the minimum value of n
required to guarantee an error of at most 1/600? (Do not find the approximation.)
Solution:
(a)(8 pts) Note that since 1 ≤ x ≤ e, we have 0 ≤ ln(x) ≤ 1 and so x2 ln(x) ≤ x2 when 1 ≤ x ≤ e. Now, we
compute the area by evaluating the integral using integration by parts with u = 1 − ln(x) ⇒ du = −dx/x and
dv = x2 dx ⇒ v = x3 /3, thus we have
Z
e
x2 (1 − ln(x))dx =
1
x3
1
(1 − ln(x)) +
3
3
Z
x2 dx
e
=
1
x3
(4 − 3 ln(x))
9
e
=
1
e3 − 4
9
(b)(8 pts) Here, since |f 00 (x)| = | − π 2 sin(πx)| ≤ π 2 , we use K = π 2 and we obtain the following error bound
|ET | ≤
K(b − a)3
K
π2
=
=⇒ |ET | ≤
.
2
12n
12 × 16
12 × 16
(c)(10 pts) How large should n be to guarantee error of at most 1/600? Using the error bound for the Midpoint
Rule, with K = 2 (since |f 00 (x)| = 2) we have
|EM | ≤
K(b − a)3
16
=⇒ |ET | ≤
.
24n2
24n2
So use n ≥ (600 × 16/24)1/2 , that is use n ≥ 20 .