 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/17/c/i/solution

u = x^{2}

d u = 2 x d x

. The convergent improper integral is

\int_{0}^{\infty} x e^{- x^{2}} d x = \frac{1}{2} \int_{0}^{\infty} e^{- u} d u = \frac{1}{2} .

 New to topics? Read the docs here!