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Solution: APPM 1360 Exam 1 Spring 2011
- (30 pts) Determine if the integrals converge or diverge. Each part is worth 10 pts.
(a)
0
ex e^2 x^ + 1 dx^ (b)
1
x √ 1 + x^6
dx (c)
1
ex^2 dx
Solution: (a) Writing the integral as a limit, we have ∫ (^) ∞
0
ex e^2 x^ + 1
dx = lim t→∞
∫ (^) t
0
ex e^2 x^ + 1
dx (^) ︸︷︷︸= u=ex t^ lim→∞ tan−^1 (et)^ −^ tan−^1 (1) =^ π/^2 −^ π/4 =^ π/^4
so the integral converges to π/4. (b) Note that 0 <
1
√^ x 1 + x^6
dx <
1
√^ x x^6
dx =
1
x^2
dx
and note that
1
x^2
dx converges, and so
1
√^ xdx 1 + x^6
converges by Direct Comparison Test.
(c) Note x^2 ≥ x for x ≥ 1 and so (^) ∫ (^) ∞
1
ex
2 ≥
1
ex^ > 0
and
1
ex^ = lim t→∞ et^ − e = ∞ and so
1
ex^2 diverges by Direct Comparison Test.
- (16 pts) Find the antiderivative. Each part is worth 8 pts.
(a)
1 − 4 x^2 dx (b)
∫ (^) cos(x) sin(x)
sin^2 (x) + 3 sin(x) + 2
dx
Solution: (a)Using the substitution x = sin(θ)/2, we have ∫ (^) √ 1 − 4 x^2 dx = (^12)
cos^2 (θ) dθ =^14
[1 + cos(2θ)] dθ
= 1 4
[
θ + sin(2θ) 2
]
=^1
[
θ + 2 sin(θ) cos(θ) 2
]
=^1
[
sin−^1 (2x) + 2x
1 − 4 x^2
]
+ C
(b) If we let u = sin(x) then du = cos(x)dx and using partial fractions yields ∫ (^) u u^2 + 3u + 2 du^ =
u + 2 du^ +
u + 1 du^ = 2 ln^ |^ sin(x) + 2| −^ ln^ |^ sin(x) + 1|^ +^ C
- (24 pts) Evaluate the given integrals. Each part is worth 8 pts.
(a)
1
xex^ dx (b)
− 1
ex −1 + ex^
dx (c)
0
x
x^2 + 1 dx
Solution:
(a) Using integration by parts with u = x and dv = exdx we get, ∫ (^2)
1
xex^ dx = xex
2
1
1
ex^ dx = ex(x − 1)
2
1
= e^2
(b) Note that the integral is undefined at x = 0, so ∫ (^1)
− 1
ex −1 + ex^
dx =
− 1
ex −1 + ex^
dx +
0
ex −1 + ex^
dx
= lim t→ 0 −
∫ (^) t
− 1
ex −1 + ex^ dx^ + lim t→ 0 +
t
ex −1 + ex^ dx {u = ex^ − 1 } → = lim t→ 0 −^
ln |et^ − 1 | − ln |e−^1 − 1 | + lim t→ 0 +^
ln |e^1 − 1 | − ln |et^ − 1 | = “∞ − ∞′′
so the integral diverges. (c) Using the u-substitution u = x^2 + 1, then du = 2xdx and so ∫ √ 3
0
x
x^2 + 1 dx =^1 2
1
u du =^1 2
u^3 /^2
4
1
3 / 2 3
(Alternate method, trigonometric substitution: Using the substitution x = tan(θ), we get ∫ x
x^2 + 1 dx =
tan(θ) sec(θ) · sec^2 (θ) dθ (^) ︸︷︷︸= u=sec(θ)
u^2 du = u
3 3 =
3 (sec(θ))
x^2 + 1)^3 3
so, ∫ √ 3
0
x
x^2 + 1 dx = (x
√ 3
0
- (a) (9 pts) What is the area of one side of a thin metal plate that covers the region above the x-axis and below the curve y = ln(x), e ≤ x ≤ e^2. (b) (6 pts) Set-up but do not solve an integral (or integrals) to find the area of the region bounded by the curves x^2 + y = 4, x + y = 2 and the lines x = −2 and x = 2. Solution:
(a) Here the area is given by
∫ (^) e 2
e
ln(x) dx, now using integration by parts with u = ln(x) and dv = dx yields (^) ∫ e^2 e
ln(x) dx = x ln(x)
e^2
e
∫ (^) e 2
e
dx = x(ln(x) − 1)
e^2
e
= e^2
(b) Here the area is given by
A =
− 2
[
(2 − x) − (4 − x^2 )
]
dx +
− 1
[
(4 − x^2 ) − (2 − x)
]
dx
