Calculus II, Winter 09 Grinshpan FINAL EXAM …tolya/122W09_final_a.pdfCalculus II, Winter 09...
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Calculus II, Winter 09 Grinshpan FINAL EXAM ANSWERS 1 A) R 2 1 1 x 3 dx = 3 8 1 B) R ln 3 0 e x √ 1+ e x dx = 4 3 (4 - √ 2) 2 A) R t 3 ln t dt = 1 4 t 4 ln t - 1 16 t 4 + C 2 B) R arctan x dx = x arctan x - 1 2 ln(x 2 + 1) + C 3 A) R sin 4 θ cos 5 θ dθ = 1 5 sin 5 θ - 2 7 sin 7 θ + 1 9 sin 9 θ + C 3 B) R dx x 2 +x-2 = 1 3 ln x-1 x+2 + C 4 A) π R 1 0 (t 2 - t 4 )dt = 2π 15 (t = x - 1) 4 B) π R 1 0 ((1 + √ y) 2 - (1 + y) 2 )dy = π 2 5 A) y = 1 2(x 2 +1) is the solution of y 0 = -4xy 2 ,y(0) = 1 2 5 B) 25πρ R 6 0 (9 - h)dh = 900πρ 6 A) 4-petal rose 6 B) dy dx θ= π 6 = tan(2θ) θ= π 6 = √ 3 7 A) R ∞ 0 √ 5e -4θ dθ = √ 5 2 7 B) 2 · 1 2 R π 2 π 6 ((3 sin θ) 2 - (1 + sin θ) 2 )dθ = π
Transcript of Calculus II, Winter 09 Grinshpan FINAL EXAM …tolya/122W09_final_a.pdfCalculus II, Winter 09...
Calculus II, Winter 09
Grinshpan
FINAL EXAM ANSWERS
1 A)∫ 2
11x3 dx = 3
8
1 B)∫ ln 3
0ex√
1 + ex dx = 43(4−
√2)
2 A)∫t3 ln t dt = 1
4t4 ln t− 1
16t4 + C
2 B)∫
arctanx dx = x arctanx− 12
ln(x2 + 1) + C
3 A)∫
sin4 θ cos5 θ dθ = 15
sin5 θ − 27
sin7 θ + 19
sin9 θ + C
3 B)∫
dxx2+x−2
= 13
ln∣∣x−1x+2
∣∣+ C
4 A) π∫ 1
0(t2 − t4)dt = 2π
15(t = x− 1)
4 B) π∫ 1
0((1 +
√y)2 − (1 + y)2)dy = π
2
5 A) y = 12(x2+1)
is the solution of y′ = −4xy2, y(0) = 12
5 B) 25πρ∫ 6
0(9− h)dh = 900πρ
6 A) 4-petal rose6 B) dy
dx
∣∣θ=π
6
= tan(2θ)∣∣θ=π
6
=√
3
7 A)∫∞
0
√5e−4θ dθ =
√5
2
7 B) 2 · 12
∫ π2
π6
((3 sin θ)2 − (1 + sin θ)2)dθ = π
