Question

$f\left(x\right)$ is a continuous and bijective function on R. If $\forall t\in R$,then the area bounded by $y=f\left(x\right),x=a-t,x=a$, and the x-axis is equal to the area bounded by $y=f\left(x\right),x=a+t,x=a$ and the x-axis. Then prove that
${\int }_{-\lambda }^{\lambda }{f}^{-1}dx=2a\lambda$ (given that $f\left(a\right)=0)$.

Answer 1

According to the question ,

$f\left(x\right)$ is bijective function and continuous

Also, $f\left(x\right)$ is symmetrical about $x=a$

i.e. $f(a-t)=f(a+t)$

Let

$y=f(a-x)a-x=f^{-1}y{\int }_{-\lambda }^{\lambda }{f}^{-1}dx={\int }_{-\lambda }^{\lambda }(a-x)dx={[ax-\frac{x^2}{2}]}_{-\lambda }^{\lambda }=[\lambda a\lambda -(-\lambda a\lambda \left)\right]-[\frac{{\lambda }^2}{2}-\frac{{\lambda }^2}{2}]=2a\lambda$

Hence the correct answer is $2a\lambda$