Answer to Question #162930 in Calculus for Pia

Question

∫x^2 sqrt(2-x)dx

Answer 1

Let $u=2-x$ , then $-du=dx$ and $x=2-u$

Use the above substitution to evaluate the integral as,

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

$=-\int (4+u^2-4u) \sqrt{u}du$

$=-\int (4\sqrt{u}+u^2\sqrt{u}-4u\sqrt{u})du$

$=-[4({\frac{2}{3}u^\frac{3}{2}})+{\frac{2}{7}u^\frac{7}{2}}-4({\frac{2}{5}u^\frac{5}{2}})]+C$

$=-\frac{8}{3}u^\frac{3}{2}-\frac{2}{7}u^\frac{7}{2}+\frac{8}{5}u^\frac{5}{2}+C$

$=-\frac{8}{3}(2-x)^\frac{3}{2}-\frac{2}{7}(2-x)^\frac{7}{2}+\frac{8}{5}(2-x)^\frac{5}{2}+C$

Therefore, the integral is $\int x^2 \sqrt{2-x}dx=-\frac{8}{3}(2-x)^\frac{3}{2}-\frac{2}{7}(2-x)^\frac{7}{2}+\frac{8}{5}(2-x)^\frac{5}{2}+C$