This is a problem from ISI MStat 2019 PSA Problem 16 based on calculating area bounded by the curve.
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 \( g^{-1} ?\)
Integration
Graph of a function
But try the problem first...
Answer: is (A)
ISI MStat 2019 PSA Problem 16
Introduction to Real Analysis by Bertle Sherbert
First hint
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 .
Second Hint
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 .
Final Step
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 .
This is a problem from ISI MStat 2019 PSA Problem 16 based on calculating area bounded by the curve.
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 \( g^{-1} ?\)
Integration
Graph of a function
But try the problem first...
Answer: is (A)
ISI MStat 2019 PSA Problem 16
Introduction to Real Analysis by Bertle Sherbert
First hint
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 .
Second Hint
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 .
Final Step
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 .
