MATH-0042 Midterm: w15exam1sol
Document Summary
Solutions to math 42 first midterm exam: (10 points) evaluate each of the following inde nite integrals. (a) r xe2xdx. Integration by parts, with u = x, v = e2x/2. We have r xe2xdx = r udv = uv r v du, so. We put x = u + 1 to make the integral r (u + 1) ln(u)du. Z (u + 1) ln(u)du = (u2/2 + u) ln(u) z (u2/2 + u) . = (u2/2 + u) ln(u) z (u/2 + 1)du = (u2/2 + u) ln(u) (u2/4 + u) + c. Finally, substituting u = x 1, we get (x2/2 1/2) ln(x 1) + ( x2/4 x/2) + c. Comment: this problem was a bit tricky, because many people tried to integrate by parts directly (rather than rst making the substitution). This led to integrate x2 x 1 , which can be done by polynomial division, but we hadn"t discussed that.
