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The area (in sq.units) of the region \left \{ \left ( x,y \right )\epsilon R^{2}\left | 4x^{2}\leq y\leq 8x+12 \right \right \} is :
Option: 1 \frac{125}{3}
Option: 2 \frac{128}{3}
Option: 3 \frac{124}{3}
Option: 4 \frac{127}{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  

 

-

 

 

\int (8x+12-4x^2)dx

Point of intersection (-1,4),(3,36)

\int_{-1}^{3}(8x+12-4x^2)dx=4x^2+12x-\frac{4x^3}{3}|_{-1}^{\;\;3}

\frac{128}{3}

Correct Option (2)

Posted by

Ritika Jonwal

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