A calculus problem by Manoj Kiran

Using integration by substitution with $\ u = x^2$ \ I = \int_{0}^{1} x^5 e^{x^2} dx = \frac{1}{2} \int_{0}^{1} u^2 e^u du = \frac{1}{2} F \........(1) Now we use integration by parts given by $\ \int t dv = tv - \int v dt$ with $\ t = u^2$ and $\ dv = e^u$ : $\ F = [u^2 e^u ]_{0}^{1} - 2 \int_0^1 u e^u du = e - 2G$ Similarily we use by parts on the integral G: $\ G = [u e^u]_0^1 - \int_0^1 e^u du = e - [e^u]_0^1 = e-(e-1) = 1$ $\therefore\ F = e -2G = e - 2$ Using (1): $\therefore\ I = \int_{0}^{1} x^5 e^{x^2} dx = \frac{1}{2} (e-2)$