Question

Let $A_n$ be the area of region bounded by a curve $y=x^3{\left(1-x^2\right)}^n,0\le x\le 1$ and the $x$-axis, then the value of $\sum _{n=1}^{\infty }{A}_n$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. 1

Answer 1

Answer: (C)

Let $A_n=\underset{0}{\overset{1}{\int }}{x}^3{\left(1-x^2\right)}^{n}dx$
Put $x^2=t\Rightarrow xdx=\frac{1}{2}dt$
$\therefore A_n=\frac{1}{2}\underset{0}{\overset{1}{\int }}t(1-t{)}^{n}dt=\frac{1}{2}\underset{0}{\overset{1}{\int }}{t}^n(1-t)dt$
$\Rightarrow A_n=\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$\therefore \sum _{n=1}^{\infty }{A}_n=\frac{1}{4}\cdot$