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 .
If the triple integral over the region bounded by the planes $2x+y+z=4$ $x=0$ $y=0$ $z=0$ is given by $\int\limits_0^2\int\limits_0^{\lambda(x)}\int\limits_0^{\mu(x,y)}\mathrm d z\mathrm d y\mathrm d x$ then the function $\lambda(x)-\mu(x,y)$ is
$x+y$
$x-y$
$x$
$y$
Key Concepts
Real Analysis
Integral Calculus
Triple Integral
Check the Answer
Answer: $\textbf{(B)} \quad y$
IIT JAM 2016, Question No. 15
Differential and Integral Calculus: R Courant
Try with Hints
Here we are given with triple integral over the region bounded by the planes $2x+y+z=4, x=0, y=0$ and $z=0$
Now we our aim here is to find $\lambda (x) $ and $\mu(x,y)$. Now we will approach this problem by find the volume of $(x,y,z)$ based on $2x+y+z=4$ can you do this ??? (With the given information x=0, y=0, z=0)
$2x+y+z=4$
$\Rightarrow z=4-2x-y$
$\Rightarrow 2x+y=4$ [as $z=0$]
$\Rightarrow y=4-2x$
Again,
$2x+y+z=4$
Now as $y=z=0$ we have $2x=4$
Therefore $x=2$
Now can you use this to move forward with this problem ?
So our triple integral become,
$\int_0^2\int_0^{(4-2x)}\int_0^{(4-2x-y)}\mathrm d z\mathrm d y\mathrm dx$
On compairing $\lambda(x)=4-2x$ and $\mu(x,y)=4-2x-y$
