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February 18, 2004 Math 112 — Exam I
Show all work clearly; an answer with no justifying computations may not receive credit (except in the “set up but do not evaluate” problems).
- (15 points) (a) Find the equation of the tangent line to the graph of x^2 + 2xy − y^2 + 4x = 5 at the point (1, 2). (b) Find (^) dxd (arctan(x/3)).
- (21 points) Find the following limits:
(a) lim x→π/ 2 −
(x −
π 2 ) tan^ x^ (b)^ xlim→ 2
x^2 − 2 x + 3 x^2 + 3 (c)^ xlim→ 1 x
1 /(x−1)
- (18 points) Set up but do not evaluate a definite integral that represents the volume of each of the following solids: (a) The base is the region bounded by the curves y = x^3 and x = y^2 , and the cross sections perpendicular to the x-axis are right triangles with height 2. (b) The solid of rotation generated by rotating about the x-axis the region above the x-axis and under y = sin x between x = 0 and x = π. (c) The solid of rotation generated by rotating the same region as in (c) about the line x = −1.
- (10 points) A liquid weighing 50 lbs/ft^3 is lifted from a cistern, from its surface to a level 40 feet above at a rate of 2 ft/sec. The vessel, which weighs 5 lb and is drawn by a rope of negligible weight, starts with 3 ft^3 of liquid, but it is leaking at a rate of .2 ft^3 /sec. Set up but do not evaluate a definite integral that represents the work (in foot-pounds) done in lifting the liquid.
- (15 points) Find the antiderivative
∫ sin 4x cos 3x dx, not by a trigonometric identity for sin mx cos nx, but by integration by parts (twice).
- (21 points) Evaluate the following integrals:
(a)
∫ sec^4 x tan^2 x dx (b)
∫ sin^4 x cos^2 x dx (c)
∫ cos(cos x) sin x dx
Some possibly useful formulas:
sin A cos B = 1 2
(sin(A + B) + sin(A − B))
sin A sin B = 1 2
(cos(A − B) − cos(A + B))
cos A cos B = 1 2
(cos(A − B) + cos(A + B))
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
∫ csc x dx = ln | csc x − cot x| + C
∫ cot x dx = ln | sin x| + C
y − y 0 = m(x − x 0 )
Subtracting the last term from both ends and dividing both ends by − 7 /9, we get ∫ sin 4x cos 3x dx = −
7 sin 4x^ sin 3x^ −^
7 cos 4x^ cos 3x^ +^ C.
- (a) Use sec^2 x = tan^2 x + 1 and the substitution u = tan x (so that du = sec^2 x dx): ∫ (tan^2 x+1) tan^2 x sec^2 x dx =
∫ (u^4 +u^2 )du =
5 u
5 +^1
3 u
3 +C =^1
5 tan
(^5) x+^1 3 tan
(^3) x+C.
(b) Use sin^2 x = 12 (1 − cos 2x) and cos^2 x = 12 (1 + cos 2x), then cos^2 2 x = 12 (1 + cos 4x), and in the last term the Pythagorean relation cos^2 2 x = 1 − sin^2 2 x and then the substitution u = sin 2x (so that du = 2 cos 2x dx): ∫ sin^4 x cos^2 x dx =
∫ (1 − cos 2x)^2 (1 + cos 2x)dx
= 1 8
∫ (1 − cos 2x − cos^2 2 x + cos^3 2 x)dx
= 1 8
( x − 1 2
sin 2x − 1 2
∫ (1 + cos 4x)dx +
∫ (1 − sin^2 2 x) cos 2x dx
)
( x − 1 2
sin 2x − 1 2
(x +^1 4
sin 4x) +^1 2
∫ (1 − u^2 )du
)
( x − 1 2
sin 2x − 1 2
(x +^1 4
sin 4x) +^1 2
(u − 1 3
u^3 )
)
= 1 8
( x − 1 2
sin 2x − 1 2
(x +^1 4
sin 4x) +^1 2
(sin 2x − 1 3
sin^3 2 x)
)
= 1 8
x − 1 8
sin 4x − 1 6
sin^3 2 x
)
(c) Let u = cos x; then du = − sin x dx, so the integral becomes
−
∫ cos u du = − sin u + C = − sin(cos x) + C.
