Browse Questions

# $I_1= \int \limits_{(-2-k)}^{k+3} x f[x (1-x)].dx; \; I_2= \int \limits_{(-2-k)}^{k+3} x f[x (1-x)].dx; \; \large\frac{I_1}{I_2}=?$

$\begin {array} {1 1} (a)\;\frac{1}{2} \\ (b)\;\frac{3}{2} \\ (c)\;\frac{1}{4} \\ (d)\;2 \end {array}$

$I_1= \int \limits_{(-2-k)}^{k+3} x f[x (1-x)].dx$
$f(x) = f(a+b-x)$
$I_1= \int \limits_{(-2-k)}^{k+3} (-2-k+k+3x) f[(-2-k+k+3)-x)[1-(-2-k+k+3)]$
$I_1= \int \limits_{-2-k}^{k+3} (-2-k+k+3x) f[(1-x)(x)].dx$
$I_1= \int \limits_{-2-k}^{k+3} (-2-k+k+3x) f[(1-x)(x)].dx- \int \limits_{-2-k}^{k+3} f[x(1-x)].dx$
$I_1=I_2-I_1$
$2I_1=I_2$
$\large\frac{I_1}{I_2}=\frac{1}{2}$
Hence a is the correct answer.