Answer to Question #273210 in Calculus for Sanjeev

Question

Prove that integral of 0 to pie √x e^-x3 dx = √pie/3

Answer 1

Indefinite integral

$$\int \sqrt{x}e^{-x^3}dx$$

Substitution $u=x^{3/2}, du=\dfrac{3}{2}x^{1/2} dx$

$$\int \sqrt{x}e^{-x^3}dx=\int u^{1/3}e^{-u^2}(\dfrac{2}{3}u^{-1/3})du$$

$$=\dfrac{2}{3}\int e^{-u^2}du$$

Use that $\displaystyle\int_{0}^{\infin}e^{-t^2}dt=\dfrac{\sqrt{\pi }}{2}$

Then

$$\displaystyle\int_{0}^{\infin}\sqrt{x}e^{-x^3}dx=\dfrac{2}{3}\displaystyle\int_{0}^{\infin}e^{-u^2}du$$

$$=\dfrac{2}{3}(\dfrac{\sqrt{\pi }}{2})=\dfrac{\sqrt{\pi }}{3}$$

Therefore

$$\displaystyle\int_{0}^{\infin}\sqrt{x}e^{-x^3}dx=\dfrac{\sqrt{\pi }}{3}$$