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

Check the Answer

Answer: is (A)

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 .

Similar Problems and Solutions

ISI MStat — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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

Check the Answer

Answer: is (A)

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 .

Similar Problems and Solutions

ISI MStat — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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