Let $R$ be the region in the first quadrant bounded by the $x$ and $y$ axis and the graphs of $f(x)=\frac{9}{25}x+b$ and $y=f^{-1}(x)$

Question

Let $R$ be the region in the first quadrant bounded by the $x$ and $y$ axis and the graphs of $f(x)=\frac{9}{25}x+b$ and $y=f^{-1}(x)$.If the area of $R$ is 49,then the value of $b$,is

$(A)\frac{18}{5}\hspace{1cm}(B)\frac{22}{5}\hspace{1cm}(C)\frac{28}{5}\hspace{1cm}(D)$none

I solved using $\int_{0}^{\frac{25b}{16}}\frac{9}{25}x+b-\frac{25}{9}(x-b)dx=49$.Where $\frac{25b}{16}$ is the point of intersection of $f(x)$ and $f^{-1}(x)$but got wrong answer.

How should i solve this problem to get correct answer?Please help.

Answer 1

Let $b>0$ :

The region is symmetric and it is enough to find the area of upper triangle :

s = $(\cfrac{25b}{16})\times(\cfrac{b}{2})$

The area of the region is $2$s

$2$s = $49$

$\cfrac{25b^{2}}{16} = 49 $

$\implies$ b = $\sqrt{\cfrac{49\times16}{25}}$

$\implies$ b = $\cfrac{28}{5}$