Question

If ${\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx=10$, then $b-a=?$

Answer 1

Step 1: Form two equations using the integral formula

It is given that ${\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx=10$

Let, $I={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx$ $...\left(\mathrm{i}\right)$

$\Rightarrow$ $I=10$

It is already known that,

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

So, on substituting $x=a+b-x$ in equation $\left(\mathrm{i}\right)$,

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

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

$\Rightarrow$ $I={\int }_a^b\frac{f\left(a+b-x\right)}{f\left(a+b-x\right)+f\left(x\right)}dx$ $...\left(\mathrm{ii}\right)$

Step 2: Calculate the required value

On adding equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we have,

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

$\Rightarrow$ $2I={\int }_a^b\left[\frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}+\frac{f\left(a+b-x\right)}{f\left(a+b-x\right)+f\left(x\right)}\right]dx$

$\Rightarrow$ $2I={\int }_a^b\left[\frac{f\left(x\right)+f\left(a+b-x\right)}{f\left(x\right)+f\left(a+b-x\right)}\right]dx$

$\Rightarrow$ $2I={\int }_a^{b}1dx$

$\Rightarrow$ $2I={\left[x\right]}_a^b$ $\mathrm{[}\mathrm{\because }\mathrm{}{\mathrm{\int }}_{\mathrm{a}}^{\mathrm{b}}\mathrm{C}\mathrm{d}\mathrm{x}\mathrm{}\mathrm{=}\mathrm{}{\mathrm{\left[}\mathrm{x}\mathrm{\right]}}_{\mathrm{a}}^{\mathrm{b}}\mathrm{]}$

$\Rightarrow$ $2I=\left[b-a\right]$

$\Rightarrow$ $2\times 10=b-a$

$\Rightarrow$ $20=b-a$

Hence, the required value of $b-a$ is $20$.