Fall 04 Midterm Answers 1 sin(tan−1 x) tan−1 x 1 π dx = √ (1 − ) 2+1 x 4 2 0 −1 u = tan x dx du = 2 x +1 Z 1. 1 Z 0 sin(tan−1 x) tan−1 x dx = x2 + 1 π/4 Z u sin u du 0 π/4 = −u cos u + sin u 0 1 π = √ (1 − ) 4 2 Z 2. x2 − x + 2 dx = 2 ln |x| − 1/2 ln(x2 + 1) − tan−1 x + C x3 + x x2 − x + 2 A Bx + C = + 2 3 x +x x x +1 x2 − x + 2 = A(x2 + 1) + (Bx + C)x A = 2 C = −1 B = −1 Z √ Z 3. e 1−x2 x Z Z Z 2 x dx x2 − x + 2 dx = dx − dx − x3 + x x x2 + 1 x2 + 1 = 2 ln |x| − 1/2 ln(x2 + 1) − tan−1 x + C dx · √ = −e 2 x 1 − x2 √ 1−x2 x +C x = sin θ dx = cos θ dθ. √ Z e 1−x2 x · dx √ = x2 1 − x2 Z Z = cos θ e sin θ · cos θ dθ sin2 θ cos θ ecot θ · csc2 θ dθ = −ecot θ + C √ = −e 1−x2 x +C Z π 4. 0 √ sin x dx x(1 + x) Converges. √ √ As x → 0, sin x ∼ x, so sin x ∼ x. √ √ sin x x 1 ∼ = 1/2 . x(1 + x) x·2 2x Z π dx converges, so by limit comparison test, the original integral 1/2 2x 0 converges. Z ∞ ln(5 + cos x) 5. dx Diverges. x2 0 Z ∞ Z 1 Z ∞ ln(5 + cos x) ln(5 + cos x) ln(5 + cos x) = + 2 2 x x x2 0 0 1 Z 1 ln 6 ln(5 + cos x) ln 6 ∼ 2 . dx diverges, so first integral diAs x → 0, 2 x2 x 0 x verges by limit comparison test. Since one part of the original integral diverges, the whole integral must diverge. ∞ X 7n ln n 6. n! n=3 Converges. n! an+1 7n+1 ln(n + 1) · n = lim n→∞ an n→∞ (n + 1)! 7 ln n 7 ln(n + 1) = lim · n→∞ n + 1 ln n =0<1 lim so series converges by Ratio Test. 7. ∞ X n=1 sin( 1 ) ln n n2 Converges. As n → ∞, 1/n2 → 0, so sin(1/n2 ) ∼ 1/n2 . ln n < n1 /2, so ∞ ∞ ∞ X X 1 ln n X n1/2 < = . 2 2 3/2 n n n n=1 n=1 n=1 The right hand series converges, so the left hand series does as well. (Direct Comparison.) Then the original series converges, by Limit Comparison. 2 8. ∞ X 89 2n + 3n+1 = n 5 30 n=2 ∞ ∞ ∞ X X 2n X 3n+1 2n + 3n+1 = + 5n 5n n=2 5n n=2 n=2 4 8 27 81 + + . . .) + ( + + . . .) 25 125 25 125 4 2 4 27 3 9 = (1 + + + . . .) + (1 + + + . . .) 25 5 25 25 5 25 4 1 27 1 = · + · 25 1 − 2/5 25 1 − 3/5 89 = 30 =( 9. ∞ X n=1 (1 + x n2 ) converges when x < 0. n Use the root test. lim a1/n = lim (1 + x/n)n n n→∞ 2 /n n→∞ = lim (1 + x/n)n n→∞ = ex . When this limit is < 1, the series converges. Thus the series converges when ex < 1, i.e. when x < 0. When this limit is > 1, the series diverges. Thus the series diverges when ex > 1, i.e. when x > 0. When the limit is equal to 1 i.e. when x = 0, the test doesn’t tell us anything, but in this case, the series is ∞ ∞ X X 2 (1)n = 1 n=1 n=1 which obviously diverges. 3