Question

Let $f$ be a positive function. If $I_1={\int }_{1-k}^{k}xfx(1-x)dx,I_2={\int }_{1-k}^{k}fx(1-x)dx,$ where $2k-1>0,$ then $\frac{I_1}{I_2}$ is

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

Answer 1

The correct option is C $\frac{1}{2}$

Using the formula:-

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I_1={\int }_{1-k}^{k}xfx(1-x)dx$ Here, $(a+b=k+1-k=1)$

$I_1={\int }_{1-k}^k(1-x)f(1-x)(1-1+x))dx$

$⟹I_1={\int }_{1-k}^k(1-x)fx(1-x)dx$

$⟹I_1={\int }_{1-k}^{k}fx(1-x)dx-{\int }_{1-k}^{k}xfx(1-x)dx$

$⟹I_1=I_2-I_1$

$⟹2I_1=I_2$

$⟹\frac{I_1}{I_2}=\frac{1}{2}$