MTH 162 Midterm: MTH 162 University of Rochester Fall 09Exam1asol

• School: University of Rochester
• Department: Mathematics
• Course: MTH 162
• Professor: All

Document Summary

October 17, 2011: (20 points) (a) (10 points) Find a partial fraction expansion for the function. 1 x3 x2 + 2x 2: (b) (10 points) Solution: (a) one notices that 1 is a root of the denominator. Polynomial division yields x3 x2 + 2x 2 = (x 1)(x2 + 2), so. 1 x3 x2 + 2x 2 (x 1)(x2 + 2) Bx + c x 1 x2 + 2. A(x2 + 2) + (bx + c)(x 1) (x 1)(x2 + 2) Ax2 + 2a + bx2 + cx bx c (a + b)x2 + (c b)x + 2a c (x 1)(x2 + 2) (x 1)(x2 + 2) By comparing numerators we must have a + b = 0, c b = 0 and 2a c = 1. From this we get a = 1/3, b = 1/3 and c = 1/3. (b) to calculate this integral, use the partial fraction expansion from (a).