Question

${\int }_0^5{x}^3(25-x^2{)}^{7/2}dx$.

Answer 1

Apply u-substitution u=$x^2$

$=\int \frac{u{(25-u)}^{\frac{7}{2}}}{2}du$

$=\frac{1}{2}\cdot \int u{(25-u)}^{\frac{7}{2}}$du

Apply integration by parts,

u=u,v'=${(25-u)}^{\frac{7}{2}}$

$=\frac{1}{2}(-\frac{2}{9}u{(25-u)}^{\frac{9}{2}}-\int -\frac{2}{9}{(25-u)}^{\frac{9}{2}}du)$

Substitute the value of u

$\frac{1}{2}(-\frac{2}{9}{x}^2{(25-x^2)}^{\frac{9}{2}}-\frac{4}{99}{(25-x^2)}^{\frac{11}{2}})$

Compute the boundaries:

${\int }_0^5{x}^3{(25-x^2)}^{\frac{7}{2}}dx=0-(-\frac{97656250}{99})=\frac{97656250}{99}$