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March 10, 2004 Math 112 — Exam II
Show all work clearly; an answer with no justifying computations may not receive credit (except for the “write down but do not evaluate” questions). If any divergent improper integrals appear, state that they are divergent.
- (8 points) Write down, but do not evaluate, an integral that represents the length of the curve y = ln | sec x| on 0 ≤ x ≤ π/4.
- (10 points) Write out the form of the partial fraction expansion of the rational function
2 x^4 + 3x^3 − x^2 + 5x + 2 x^4 + 3x^2
You need not find the constants in the numerators, but be sure to complete the first step.
- (10 points) A tank with the dimensions in the figure is filled to a height of 3 ft with a liquid weighing 60 lb/ft^3. Write down, but do not evaluate, an integral that represents the total force, in pounds, exerted by the liquid on the triangular door in the front of the tank. (It may be that not all of the data given is needed.)
- (7 points) Determine whether the improper integral ∫ (^) ∞
1
arctan x 1 + x^2 dx
is convergent or divergent. Explain your answer.
- (65 points) Evaluate the following integrals:
(a)
∫ (^2)
0
3 x + 2 √ x^2 + 4
dx (b)
∫ (^3)
0
3 x + 2 x^2 − 4
dx
(c)
∫ x √ 8 + 2x − x^2
dx (d)
∫ (^) ∞
0
x^2 e−xdx
Some possibly useful formulas:
sin^2 θ + cos^2 θ = 1 tan^2 θ + 1 = sec^2 θ sec^2 θ − 1 = tan^2 θ
sin 2A = 2 sin A cos A cos 2A = cos^2 A − sin^2 A = 2 cos^2 A − 1 = 1 − 2 sin^2 A
∫ tan x dx = ln | sec x| + C
∫ sec x dx = ln | sec x + tan x| + C
∫ u dv = uv −
∫ v du (x, y) =
( ∫ x(f (x) − g(x))dx area
∫ (^1) 2 (f^ (x)
(^2) − g(x) (^2) )dx area
)
t = tan x 2
: cos x = 1 − t^2 1 + t^2
, sin x = 2 t 1 + t^2
, dx =
1 + t^2
dt
(b) First, we should note that the denominator is 0 at 2, inside the interval, so the integral is improper and may be divergent — we will need to find the limit of the antiderivative as x approaches 2 from both sides. We need to find the antiderivative: We factor the denominator: (x−2)(x+2); so the form of the partial fraction expansion of the integrand is (3x + 2)/(x^2 − 4) = A/(x − 2) + B/(x + 2). Multiplying both sides by x^2 − 4 gives 3 x + 2 = A(x + 2) + B(x − 2) = (A + B)x + (2A − 2 B), or A + B = 3 and 2A − 2 B = 2. Thus, A = 2 and B = 1, so ∫ (^3) x + 2 x^2 − 4 dx =
x − 2
x + 2
) dx = 2 ln |x − 2 | + ln |x + 2| + C.
Now the limit as x approaches 2 (from either direction) of the first term is unbounded, so the improper integral is divergent. (c) Because 8 + 2x − x^2 = 8 − (x^2 − 2 x + 1) + 1 = 9 − (x − 1)^2 , we set x − 1 = 3 sin θ, so that x = 3 sin θ + 1, dx = 3 cos θ dθ, and
8 + 2x − x^2 = 3 cos θ. So the integral becomes ∫ (^) 3 sin θ + 1 3 cos θ 3 cos θ dθ =
∫ (3 sin θ + 1) dθ = −3 cos θ + θ + C
= −
√ 8 + 2x − x^2 + arcsin
( (^) x − 1 3
)
(d) We will need to use integration by parts twice, first with u = x^2 and dv = e−xdx, then with u = x and dv = e−xdx: ∫ (^) ∞
0
x^2 e−xdx = lim t→∞
∫ (^) t
0
x^2 e−xdx = lim t→∞
[ −x^2 e−x^ + 2
∫ xe−xdx
]t
0 = lim t→∞
[ −x^2 e−x^ − 2 xe−x^ + 2
∫ e−xdx
]t
0 = lim t→∞
[ −x^2 e−x^ − 2 xe−x^ − 2 e−x
]t 0
=
(
t^ lim→∞
−t^2 − 2 t − 2 et
)
=^ H
( lim t→∞
− 2 t − 2 et
)
( lim t→∞
et
)
