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 $y=f^{-1}(x)$
By Abigail Rogers •
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.
$\endgroup$1 Answer
$\begingroup$Let b>0 :
The region is symmetric and it is enough to find the area of upper triangle :
s=(25b/16)(b)/2
The area of the region is 2s
2s=49
25b^2/16 = 49 : b = 28/5
$\endgroup$
