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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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Let f:RrarrR be a continuous and bijective function defined such that f(alpha)=0(alpha ne0) . The area bounded by y=f(x),x=alpha,x=alpha-t is equal to the area bounded by y=f(x),x=alpha,x=alpha+tAAtin R then Graph of y=f(x) is symmetrical about point