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# ISI MStat 2019 PSA Problem 16 | Area bounded by the curve

This is a problem from ISI MStat 2019 PSA Problem 16 based on calculating area bounded by the curve.

## Area bounded by the curve - ISI MStat 2018 PSA Problem 16

The functions $$f, g:[0,1] \rightarrow[0,1]$$ are given by $$f(x)=\frac{1}{2} x(x+1)$$ and $$g(x)=\frac{1}{2} x^{2}(x+1) .$$ What is the area enclosed between the graphs of $$f^{-1}$$ and $$g^{-1} ?$$

• (A) 1/8
• (B) 1/4
• (C)5/12
• (D) 7/24

### Key Concepts

Integration

Graph of a function

ISI MStat 2019 PSA Problem 16

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Inverse of a function is basically reflection about y=x line .

So , we can get $$f^{-1}$$ and $$g^{-1}$$ from $$f(x)$$ and $$g(x)$$ respectively by replacing x by y and y by x .

Let's draw the curves .

This is graph of inverses of f and g when they are defined in $$R \to R$$ . But in our problem we should consider one positive x axis and y axis .

Therefore area of the curve bounded by the graphs of $$f^{-1}$$ and $$g^{-1}$$ is $$\int^1_{0}\frac{y^2+y-y^3-y^2}{2} \,dx$$ = 1/8 .