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  • Given : <img alt="f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right." src="https://learn.careers360.com/latex-image/?f%28x%29%3D%5Cleft%5C%7B%5Cbe

Given : f(x)=\left\{\begin{matrix} x, &0\leq x< \frac{1}{2} \\ \frac{1}{2}, &x=\frac{1}{2} \\ 1-x, &\frac{1}{2} < x\leq 1 \end{matrix}\right.  and g(x)=\left ( x-\frac{1}{2} \right )^{2},\: x\epsilon \textbf{R}. Then the area (in sq. units) of the region bounded b the curves, y=f(x)  and y=g(x) between the lines, 2x=1\: \: \: and\: \: \: 2x=\sqrt{3}, is : 
Option: 1 \frac{\sqrt{3}}{4}-\frac{1}{3}
 
Option: 2 \frac{1}{3}+\frac{\sqrt{3}}{4}
 
Option: 3 \frac{1}{2}+\frac{\sqrt{3}}{4}
 
Option: 4 \frac{1}{2}-\frac{\sqrt{3}}{4}
 
 

Answers (1)

best_answer

 

 

Area Bounded by Curves When Intersects at More Than One Point -

Area bounded by the curves  y = f(x),  y = g(x)  and  intersect each other in the interval [a, b]

First find the point of intersection of these curves  y = f(x) and  y = g(x) , let the point of intersection be x = c

Area of the shaded region  

=\int_{a}^{c}\{f(x)-g(x)\} d x+\int_{c}^{b}\{g(x)-f(x)\} d x

 

When two curves intersects more than one point

rea bounded by the curves  y=f(x),  y=g(x)  and  intersect each other at three points at  x = a, x = b amd x = c.

To find the point of intersection, solve f(x) = g(x).

For x ∈ (a, c), f(x) > g(x) and for x ∈ (c, b), g(x) > f(x).

Area bounded by curves,

\\\mathrm{A=} \int_{a}^{b}\left |f(x)-g(x) \right |dx\\\\\mathrm{\;\;\;\;=} \int_{a}^{c}\left ( f(x)-g(x) \right )dx+\int_{c}^{b}\left ( g(x)-f(x) \right )dx  

 

-

 

 

Required area = Area of trapezium ABCD - \int_{1/2}^{\sqrt3/2}\left ( x-\frac{1}{2} \right )^2dx

\\=\frac{1}{2}\left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\left(x-\frac{1}{2}\right)^{3}\right)_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\\=\frac{\sqrt3}{4}-\frac{1}{3}

Correct Option (1)

Posted by

avinash.dongre

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