Summer 2006 Final Solutions 1. (a) 4 Z dx 1 = 16 + x2 16 0 Let u = x 4 dx 4 . so that du = 1 4 Z 1 0 Z 4 dx 1 + ( x4 )2 0 This gives du 1 = (arctan 1 − arctan 0) 1 + u2 4 π 16 = (b) Pick u = x2 so that du = 2x dx. Now Z (x2 ) x3 1 dx = 2 Z 3u du = 3u +C 2 ln 3 2 = 3(x ) +C 2 ln 3 (c) This was done in class. (d) Pick u = ln x so that du = Z 1 e dx x . Now (ln x)2 dx = x = Z 1 u2 du 0 1 3 (e) Integrate by parts twice. Pick u = ex so that du = ex dx. Pick dv = sin x dx so that v = − cos x. Now Z ex sin x dx = −ex cos x + Z ex cos x dx Pick now u = ex and dv = cos x dx so that v = sin x. Now we have Z ex sin x dx = −ex cos x + ex sin x − Z ex sin x dx Add the integral on the right to both sides and divide by 2 to get Z ex sin x dx = 1 1 x (e sin x − ex cos x) 2 R 2. (a) v(t) = a(t) dt = −6t + C1 . To solve for C1 notice that 3 = v(0) = −6(0) + C1 = C1 so that C1 = 3. Plug this in to get v(t) = −6t + 3 (b) x(t) = v(t) dt = −3t2 + 3t + C2 . To solve for C2 notice that 0 = x(0) = C2 . Plug this in to get R x(t) = −3t2 + 3t (c) x(1) = 0 3 ≥ 0 iff t ≤ 1 2 R1 |v(t)| dt = R1 | − 6t + 3| dt. Notice that −6t + R1 R1 so that this integral turns into 02 3 − 6t dt + 1 6t − 3 dt. (d) Distance traveled is 0 Evaluate these integrals to get 3. (a) F (0) = 0 and F (4) = 3 2 0 2 units. 4 3 (b) F 0 (x) = x2 − 4x + 3 = (x − 3)(x − 1). This is negative for 1 < x < 3 (F is decreasing here), and positive for x < 1 and for x > 3 (F is increasing here). 1 and 3 are critical points, the first being a local max and the second being a local min. (c) F 00 (x) = 2x − 4 This is negative for x < 2 (F is concave down here), and positive for x > 2 (F is concave up here). Thus x = 2 is an inflection point for F . (d) F (1) = 23 , F (2) = 23 , and F (3) = 0. 4. (b) Uf (P) = 31 and Lf (P) = 7. 5. (a) To find the intersection points, set y1 = y2 : x2 + 1 = 2x + 1 x2 − 2x = 0 x(x − 2) = 0 so that the intersection points are 0 and 2. To find which function is the upper bounding function, plug x = 1 into both and get y1 (1) = 2 and y2 (1) = 3. Thus y2 is the upper bounding function. The area between the curves is Z 2 Z y2 − y1 dx = 0 2 2x − x2 dx 0 =4− 2 8 4 = 3 3