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f(x) is a continuous and bijective funct...

`f(x)` is a continuous and bijective function on `Rdot` If `AAt in R ,` then the area bounded by `y=f(x),x=a-t ,x=a ,` and the x-axis is equal to the area bounded by `y=f(x),x=a+t ,x=a ,` and the x-axis. Then prove that `int_(-lambda)^lambdaf^(-1)(x)dx=2alambda(gi v e nt h a tf(a)=0)dot`

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Given `|int_(a-t)^(a)f(x)dx|=|int_(a)^(a+t)f(x)dx|AAtepsilonR`
or `int_(a-t)^(a)f(x)dx=-int_(a)^(a+t)f(x)dx`
[since `f(a)=0` and `f(x)` is monotonic]
or `f(a-t)=-f(a-t)` (differentiating both side w.r.t `t`)
or `f(a+t)=-f(a-t)=x` (say)
or `t=f^(-1)(x)-a`..........2
and `t=a-f^(-1)(-x)` ............3
From equations 3 and 2 `(a-f^(-1)(x))+(a-f^(1)(-x))=0`
or `int_(-lamda)^(lamda) f^(-1)(x)dx=1/2 int_(-lamda)^(lamda)(f^(-1)(x)+f^(-1)(-x))dx=2alamda`
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