advertisement

MATH 409.501 Spring 2009 Final Exam Solutions 2. (b) 4. (d) 5. (c) 8. (a) F 0 (x) = 2xex 4 (b) Using the substitution u = −x we see that Z 0 Z 1 4x 4x − dx = dx 2 2 −1 x + 3 0 x +3 so that Z 1 −1 Z 0 Z 1 4|x| 4x 4x dx = − dx + dx 2 2 x2 + 3 −1 x + 3 0 x +3 Z 1 1 4 4x 2 dx = 4 ln(x + 3) = 4 ln . =2 2 3 0 0 x +3 e−x ≤ e−x . Also, for b ≥ 0 we have x2 + 3 b Z b −x −x e dx = −e = −e−b + 1 (c) For x ∈ [0, ∞) we have 0 ≤ 0 0 and this last quantity converges to 1 as b → ∞. Thus by the comparison theorem f is improperly integrable on (0, ∞). 11. (b) Let ε > 0. Take an n ∈ N such that n > 2/ε. Then for the partition P = {0, n1 , n2 , n3 , . . . , n−1 , 1} of [0, 1] we have n U (f, P ) = 1 · 1 2 1 3 1 n 1 1 1 + · + · + · · · + · = + 2 (2 + 3 + · · · + n) n n n n n n n n n and L(f, P ) = 0 · 1 1 1 2 1 n−1 1 1 + · + · + ··· + · = 2 (1 + 2 + · · · + (n − 1)) n n n n n n n n so that U (f, P ) − L(f, P ) = Thus f is integrable on [0, 1]. 1 n 1 2 1 2 + 2 − 2 = − 2 < < ε. n n n n n n