Question

Let $J={\int }_0^{\frac{1}{2}}(\frac{1}{4}-x^2{)}^{4}dxandK={\int }_0^{\frac{1}{2}}{x}^4(1-x{)}^{4}dx,$ then

• A. $\frac{J}{K}=2$
• B. $J-K=0$
• C. $J=\frac{1}{2}{\int }_0^1{x}^4(1-x{)}^{4}dx$
• D. $K=\frac{1}{1260}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $J-K=0$
C $J=\frac{1}{2}{\int }_0^1{x}^4(1-x{)}^{4}dx$
D $K=\frac{1}{1260}$

$J={\int }_0^{\frac{1}{2}}(\frac{1}{4}-x^2{)}^{4}dx={\int }_0^{\frac{1}{2}}(\frac{1}{4}-(\frac{1}{2}-x{)}^2{)}^{4}dxK={\int }_0^{\frac{1}{2}}(x-x^2{)}^{4}dx={\int }_0^{\frac{1}{2}}{x}^4(1-x^4)dx...\left(i\right)J=K,soJ-K=0Putx=1-yK=-{\int }_1^{\frac{1}{2}}(1-y{)}^2{y}^{4}dyK={\int }_{\frac{1}{2}}^1{x}^4(1-x{)}^{4}dx...\left(ii\right)equation\left(i\right)+\left(ii\right)\Rightarrow 2J={\int }_0^1{x}^4(1-x^{4}dx)$
$J=\frac{1}{2}{\int }_0^1{x}^4(1-x{)}^{4}dxJ=\frac{1}{2}{\int }_0^1{x}^4(1-x^4)dxPutx=si{n}^2\theta J={\int }_0^{\frac{\pi }{2}}si{n}^9\theta .co{s}^9\theta d\theta =\frac{(8\times 6\times 4\times 2)(8\times 6\times 4\times 2)}{(18\times 16\times 14\times 12\times 10\times 8\times 6\times 4\times 2)}=\frac{1}{1260}$