Question

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

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

Answer 1

Answer: (C)

Let $A_n=\int\limits_0^1 x^3\left(1-x^2\right)^n d x$
Put $\quad x^2=t \Rightarrow x d x=\frac{1}{2} d t$
$\therefore A_n=\frac{1}{2} \int\limits_0^1 t(1-t)^n d t=\frac{1}{2} \int\limits_0^1 t^n(1-t) d t $
$\Rightarrow A_n=\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$
$\therefore \displaystyle\sum_{n=1}^{\infty} A_n=\frac{1}{4} \cdot$