Question

Let $J=\underset{0}{\overset{1/2}{\int }}{(\frac{1}{4}-x^2)}^{4}dx$ and $K=\underset{0}{\overset{1/2}{\int }}{x}^4(1-x{)}^{4}dx.$ Then

• A. $K=\frac{1}{1260}$
• B. $J=\frac{1}{2}\underset{0}{\overset{1}{\int }}{x}^4(1-x{)}^{4}dx$
• C. $J-K=0$
• D. $\frac{J}{K}=2$

Answer 1

Answer: (A), (B), (C)

$J-K=0$

Given: $J=\underset{0}{\overset{1/2}{\int }}{(\frac{1}{4}-x^2)}^{4}dx$
We know that $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\therefore J=\underset{0}{\overset{1/2}{\int }}{(\frac{1}{4}-{(\frac{1}{2}-x)}^2)}^{4}dx$
$\Rightarrow J=\underset{0}{\overset{1/2}{\int }}{(x-x^2)}^{4}dx$
$\Rightarrow J=\underset{0}{\overset{1/2}{\int }}{x}^4(1-x{)}^{4}dx=K$
$\therefore J-K=0$ and $\frac{J}{K}=1$


We have, $J=\frac{2}{2}\underset{0}{\overset{1/2}{\int }}{x}^4(1-x{)}^{4}dx$
We know that $\underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$ if $f(2a-x)=f\left(x\right).$
$\therefore J=\frac{1}{2}\underset{0}{\overset{1}{\int }}{x}^4(1-x{)}^{4}dx=K$


We have,
$K=\frac{1}{2}\underset{0}{\overset{1}{\int }}{x}^4(1-x{)}^{4}dx$
Put $x={\sin }^2\theta \Rightarrow dx=2\sin \theta \cos \theta d\theta$
$\therefore K=\underset{0}{\overset{\pi /2}{\int }}{\sin }^9\theta \cdot {\cos }^9\theta d\theta$
$\Rightarrow K=\frac{(8\times 6\times 4\times 2)(8\times 6\times 4\times 2)}{(18\times 16\times 14\times 12\times 10\times 8\times 6\times 4\times 2)}$
$\Rightarrow K=\frac{(8\times 6\times 4\times 2)}{(18\times 16\times 14\times 12\times 10)}$
$\therefore K=\frac{1}{1260}$