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The area (in sq.units) of the region \left \{ (x,y) \epsilon R^{2}:x^{2}\leq y\leq 3-2x\right \}, is :
Option: 1 \frac{31}{3}

Option: 2 \frac{32}{3}

Option: 3 \frac{29}{3}

Option: 4 \frac{34}{3}
 

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  

 

-

 

 

Point of intersection of y=x^{2}\;\; \&\;\; y=-2 x+3 is – 3, 1

Now, \text { Area } 3:=\int_{-3}^{1}\left(-2 x+3-x^{2}\right) d x

\\\left[- x^2+3x-\frac{x^{3}}{3}\right]^{\;\;\;1}_{-3}=-2\left(\frac{1^{2}-3^{2}}{2}\right)+3(1-(-3))-\left(\frac{1^{3}+3^{3}}{3}\right)

=12+8-\frac{28}{3}=\frac{32}{3}

Correct Option (2)

Posted by

vishal kumar

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