MATH-0042 Midterm: w14exam2sol
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Solutions to math 42 second exam february 20, 2014: (12 points) (a) evaluate z . 0 xe x2 dx or explain why its value does not exist; show all reasoning. (5 points) the function f (x) = xe x2. I and we should compute is continuous at 0, so we have an improper integral of type lim t z t. We use u-substitution: let u = x2 so that du = 2x dx. The bounds of integration become u(0) = 0 and u(t) = t2, hence we obtain lim t z t2. 0 ln(1 + x) x2 dx converges or diverges; give complete reasoning. (7 points) the function f (x) = ln(1+x) continuous at 0 so we have an improper integral of type ii. = lim x 0+ ln(1 + x) x2 x. The last fraction is an indeterminate form of type 0 and positive when we take the limit for x 0+, hence using l"h opital"s rule we obtain.
