# The area of the plane region bounded by the curves $\dpi{100} x+2y^{2}=0\; and\; x+3y^{2}=1$  is equal to Option 1) $\frac{4}{3}\;$ Option 2) $\; \frac{5}{3}\;$ Option 3) $\; \frac{1}{3}\;$ Option 4) $\; \frac{2}{3}$

As learnt in concept

Area along y axis -

Let $y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x)$ be two curve, then area bounded by the curves and the lines

y = a and y = b is

$A=\int_{a}^{b}\left ( x_{2}-x_{1} \right )dy$

- wherein

x + 2y2=0 and x+3y2= 1

=>$2\int_{0}^{+1}(1-3y^{2})-(-2y^{2})dy$

=$2\int_{-0}^{+1}(1-y^{2})dy$

=$2\left [ y-\frac{y^{2}}{3} \right ]_{0}^{1}$

$2\times (1-\frac{1}{3})=\frac{4}{3}$

Option 1)

$\frac{4}{3}\;$

this is correct option

Option 2)

$\; \frac{5}{3}\;$

this is incorrect option

Option 3)

$\; \frac{1}{3}\;$

this is incorrect option

Option 4)

$\; \frac{2}{3}$

this is incorrect option

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