Question

Calculate ${\int }_0^1{x}^2\sqrt{1-x}dx$ using reduction formulae.

• A. $-\frac{4}{105}$
• B. $\frac{4}{105}$
• C. $-\frac{16}{105}$
• D. $\frac{16}{105}$

Answer 1

The correct option is D $\frac{16}{105}$

We have to find ${\int }_0^1{x}^2\sqrt{1-x}dx$ using reduction formula.

Let $I_n={\int }_0^1{x}^n\sqrt{1-x}dx$

Take $u=x^n,dv=\sqrt{1-x}dx$

$\Rightarrow du=n{x}^{n-1}dx,v=-\frac{2}{3}(1-x{)}^{3/2}$

$I_n={[-\frac{2}{3}{x}^n(1-x{)}^{3/2}]}_0^1+\frac{2n}{3}{\int }_0^1\left\{\right(1-x{)}^{3/2}{x}^{n-1}\}dx$

$=[0-0]+\frac{2n}{3}{\int }_0^1\left\{\right(1-x\left)x^{n-1}\sqrt{1-x}\right\}dx$

$=\frac{2n}{3}{\int }_0^1\left\{\right(x^{n-1}-x^n\left)\sqrt{1-x}\right\}dx$

$=\frac{2n}{3}{\int }_0^1\{x^{n-1}\sqrt{1-x}-x^n\sqrt{1-x}\}dx$

$I_n=\frac{2n}{3}{\int }_0^1\left\{x^{n-1}\sqrt{1-x}\right\}dx-\frac{2n}{3}{\int }_0^1\left\{x^n\sqrt{1-x}\right\}dx$

$\therefore I_n=\frac{2n}{3}{I}_{n-1}-\frac{2n}{3}{I}_n$

$\Rightarrow 3I_n=2n{I}_{n-1}-2n{I}_n$

$\Rightarrow (3+2n)I_n=2n{I}_{n-1}$

$\therefore I_n=\frac{2n}{3+2n}{I}_{n-1}...\left(1\right)$

We have to find $I_2={\int }_0^1{x}^2\sqrt{1-x}dx$

Let us find $I_0={\int }_0^1{x}^0\sqrt{1-x}dx={\int }_0^1\sqrt{1-x}dx={[-\frac{2}{3}(1-x{)}^{3/2}]}_0^1=\frac{2}{3}$

$I_1={\int }_0^{1}x\sqrt{1-x}dx$

By $\left(1\right)$, $I_1=\frac{2}{5}{I}_0=\frac{2}{5}\times \frac{2}{3}=\frac{4}{15}$

$I_2={\int }_0^1{x}^2\sqrt{1-x}dx$

By $\left(1\right)$, $I_2=\frac{4}{7}{I}_1=\frac{4}{7}\times \frac{4}{15}=\frac{16}{105}$

Thus ${\int }_0^1{x}^2\sqrt{1-x}dx=\frac{16}{105}$