Math 12, Fall 2006 Solutions to the Final Exam

Math 12, Fall 2006 4 Solutions to the Final Exam Z 2 (x = 3 tan θ, dx = 3 sec θ dθ) = Z ∗∗ dx 1 ∗∗ 3 sec2 θ dθ 1(a). = = cos θ dθ 2 3/2 27 sec3 θ 9 ∗ 0 (9 + x ) ∗ 4 ∗∗ 4 1 x 4 1 √ = sin θ = · = √ −0= , 9 9 45 9 + x2 0 9 25 ∗ where the third-to-last equality is by a right triangle with angle θ, height x, base 3, and therefore √ hypoteneuse 9 + x2 . Z Z du dx −1 √ p = arcsin u + C = (u = ln x, du = x dx) = 1(b). 1 − u2 x 1 − (ln x)2 = arcsin(ln x) + C. Z 12 1(c). Applying partial fractions to dx, we rewrite the integrand as A/x + B/x2 + 2 x (x − 2) C/(x − 2). Adding those fractions and comparing the numerator to the original numerator of 12, we get Ax(x − 2) + B(x − 2) + Cx2 = 12, i.e., (A + C)x2 + (B − 2A)x − 2B = 12. Either by evaluating the first equality at x = 0 and x = 2 and, say, x = 1, or by solving the three equations A+C = 0, B −2A = 0, −2B = 12 given by the second equality, we get A = −3, B = −6, C = 3. SoZ the integral is Z 12 −3 6 3 6 dx = − + dx = + 3 ln |x − 2| − 3 ln |x| + C. x2 (x − 2) x x2 x − 2 x Z 0 Z 0 dx dx 2.(a) Write = lim because the integral is improper at −∞ and 2 t→−∞ t 1 + 4x2 −∞ 1 + 4x nowhere else. Z Z 0 1 0 du dx = (u = 2x, du = 2 dx) = The integral with t is 2 2 2t 1 + u2 t 1 + 4x 0 1 1 = arctan u = 0 − arctan(2t). 2 2 2t Z 0 dx 1 1  π π So the original integral is = lim − arctan(2t) = − − = . 2 t→−∞ 2 2 2 4 −∞ 1 + 4x Z 0 Z 3 Z 3 dx dx dx = + , since the integral is improper at 0 (in the middle). 2(b). Write 4 4 4 −1 x 0 x −1 x Z 0 Z t t 1 1 1 dx dx = lim− = lim− − 3 The first of these two integrals is = lim− − 3 − = ∞. 4 4 t→0 t→0 3x −1 t→0 3t 3 −1 x −1 x Thus, this integral, and hence the original integral also, diverges. Z sinh x cosh x 0 0 1 cosh x − 1 = L’H = lim = L’H = lim = . 2 x→0 2x x→0 x→0 x 0 0 2 2 3(a). lim 1/x2 3(b). Write lim+ (cos x) x→0  ln(cos x) = exp lim+ x→0 x2  because the original limit is a 1∞ indeter- minate form. − sin x cos x = lim − sin x · cos x x→0+ 2x x 2 0 ln(cos x) = L’H = lim+ 2 x→0 x→0 x 0 1 1 −1/2 . = −1 · = − . So the original limit is exp(−1/2) = e 2 2 ∞ X X 1 sin2 (2n) sin2 (2n) 1 4(a). For the series , note that 0 ≤ ≤ . Since converges by the 2n 2n 2n 2n n=2 Geometric Series Test, the original series therefore converges by the Comparison Test. The limit inside the exponential is lim+ 4(b). Applying the Ratio Test to ∞ X (2n)! n=1 nn , we get (2n + 2)!  n (2n + 1)(2n + 2) n 2(2n + 1) 2·∞ (n + 1)n+1
n = lim · = ∞ > 1, = lim = lim 1 n→∞ n→∞ n→∞ 1 + (2n)! n+1 n+1 e n nn so the series diverges by the Ratio Test. √ n n−1/2 is alternating, and we check lim = lim = 0. 5. The series n→∞ n + 68 n→∞ 1 + 68n−1 n + 68 n=0 √ x Moreover, the terms decrease for n ≥ 68, because if we let f (x) = , then x + 68 √ 68 − x (1/2)x−1/2 (x + 68) − x = √ f ′ (x) = <0 for x > 68. 2 (x + 68) 2 x(x + 68)2 Thus, the series converges by the Alternating Series Test. √ ∞ ∞ X X n 1 to by LCT: Meanwhile, we can compare the absolute series 1/2 n + 68 n n=1 n=0 √ n 1 n = lim = 1 > 0, lim n + 68 = lim 1 n→∞ 1 + 68n−1 n→∞ n + 68 n→∞ n1/2 ∞ X 1 so the two series are comparable. Since diverges by the p-test, the original absolute 1/2 n n=1 series diverges by the Limit Comparison Test. Thus, the original series converges conditionally. √ ∞ X (−1)n n 6(a). ∞ X (−1)n · 2n+2 is a geometric series with initial term a = −8/3 and ratio r = −2/3. 3n Thus, the sum is a/(1 − r) = (−8/3)/(5/3) = −8/5. n=1 ∞ ∞ X X 1 (−1)n (π/6)2n+1 (−1)n π 2n+1 = = sin(π/6) = . 6(b). 2n+1 6 (2n + 1)! n=0 (2n + 1)! 2 n=0 2 7. Applying the Ratio Test to ∞ X (x + 2)n n=1 n · 3n gives (x + 2)n+1 1 |x + 2| |x + 2| |x + 2| n (n + 1)3n+1 · = lim · = lim = lim n −1 n→∞ 1 + n n→∞ n→∞ n + 1 (x + 2) 3 3 3 n n·3 The series converges if this limit is < 1 and diverges if the limit is > 1. Expanding |x + 2|/3 < 1 gives −3 < x + 2 < 3, i.e., −5 < x < 1. X 3n X1 Checking the endpoints, observe that at x = 1 the series is , which diverges = n n3 n X (−1)n , which is alternating, and the absolute terms by the p-test. At x = −5, the series is n 1/n decrease with limit 0; so the series converges by the Alternating Series Test. Thus, the interval of convergence is [−5, 1). x4 x 8(a). To find the Maclaurin series for g(x) = 3xe , start with e = x4 e = ∞ X x4n n=0 n! , and therefre the desired series is g(x) = 3xe x4 ∞ X xn n! , so that n=0 4n+1 = ∞ X 3x n=0 n! . ∞ X 1 x , start with = xn and 8(b). To find the Maclaurin series for f (x) = (1 + 2x)2 1−x n=0 ∞ X 1 differentiate to get = nxn−1 . Substituting −2x for x gives (1 − x)2 n=1 ∞ X 1 = n · (−2)n−1 xn−1 , so that the desired series is 2 (1 + 2x) n=1 ∞ X x = n · (−2)n−1 xn . f (x) = 2 (1 + 2x) n=1 9(a). (See me for a sketch of the region R beneath the graph of y = ex and above the x-axis, for 0 ≤ x ≤ 2). Slice the solid formed by revolving R about the line y = −1 with vertical slices. The slice at position x is an annulus with inner radius 0−(−1) = 1, and with outer radius ex −(−1) = ex +1. The slices run from x = 0 to x = 2. Thus, the volume  is 2 Z 2 Z 2 1 2x π 2x x x 2 2 x π(e + 2e ) dx = π e + 2e π(e + 1) − π(1) dx = = (e4 + 4e2 − 5). 2 2 0 0 0 9(b). Slice the solid formed by revolving R about the line x = 4 into vertical cylinders. The slice passing through position x has radius 4 − x and height ex ; the outermost cylinder is at x = 0 and the innermost is at x = 2. Thus, the volume is Z 2 0 2π(4 − x)ex dx = (u = 4 − x, dv = 2πex dx, du = −dx, v = 2πex ) Z 2 h i2 2 x x 2 = 2π(4 − x)e + 2πe dx = 2π(2e − 4) + 2π ex = 2π(3e2 − 5). 0 0 0 3 10(a). For the parametric curve C given by x = t + t2 , y = t − t2 for 0 ≤ t ≤ 2, we compute 1 − 2t dy = . x′ (t) = 1 + 2t and y ′ (t) = 1 − 2t, so that dx 1 + 2t 10(b). See me for a sketch. Roughly speaking, since dy/dx is positive for t < 1/2 and negative for t > 1/2 (hitting zero at t = 1/2 but never hitting ∞), and since (x, y) is (0, 0) for t = 0, is (3/4, 1/4) for t = 1/2, and is (6, −2) for t = 2, we can describe it as follows. The curve starts from the origin, slopes up to the upper right towards a peak at (3/4, 1/4), then slopes down to the lower right to the point (6, −2). Besides starting at (0, 0), the curve never again hits the y-axis, since x(t) > 0 for t > 0. However, since y = 0 when t = 0, 1, and since x(1) = 2, the curve crosses the x-axis on the way down at the point (2, 0). 10(c). Since (x′ (t))2 + (y ′ (t))2 = (1 + 2t)2 + (1 − 2t)2 = 2 + 8t2 , the integral for the length of C Z 2√ is 2 + 8t2 dt. 0 10(d). Similarly, the integral for the surface area generated by revolving C about the x-axis is Z 2 √ 2π(t − t2 ) 2 + 8t2 dt. 0 11(a). See me for a sketch the curve r = sin(3θ). (It is a 3-leaf rose, with one petal pointing down, and the other two pointing into the first and second quadrants, at angles π/6 and 5π/6. See, for example, the book’s solution to 11.4 #13 (in the answer appendix, page A95), but rotated 90 degrees clockwise.) 11(b). To find the area enclosed, it’s important to realize that the curve is traced out in full as θ runs only from 0 toZπ. Thus, the area enclosed is Z π π 1 π 1 π θ π 1 2 sin (3θ) dθ = sin(6θ) = − 0 − 0 + 0 = . 1 − cos(6θ) dθ = − 4 0 4 24 4 4 0 2 0 4