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Question

The area bounded by the curves $$ x= y^{2} $$ and $$ x=3-2y^{2} $$ is:


Solution

From figure, the two curves represent  parabolas with vertices at $$ (0,0)$$ and $$(3,0)$$. They intersect at $$(1,1)$$ and $$(1,-1),$$ so the required area is 
here,
$$x=y^2$$  $$\Rightarrow$$ $$y=\sqrt x$$
also
$$x=3-2y^2$$  $$\Rightarrow$$ $$-2y^2=x-3$$$$\Rightarrow$$ $$y=\sqrt {\dfrac{3-x}{2}}$$

$$\Rightarrow$$ area of $$OPMQO = 2 $$ (area of $$OPMO)$$

$$\displaystyle =2\left( \int_{0}^{1} \sqrt{x} \: dx +\int_{1}^{3} \sqrt{\frac{3-x}{2}} dx \right)$$
$$\displaystyle =\left. \left( \frac{2}{3} x^{3/2} \right| _{0}^{1} \left. -\frac{1}{\sqrt{2}} \cdot \frac{2}{3} (3-x)^{3/2} \right|_{1}^{3} \right) $$
$$\displaystyle =2 \left[ \frac{2}{3}-\left( 0-\frac{1}{\sqrt{2}} \cdot \frac{2}{3} 2^{3/2}\right) \right] =2 \left( \frac{2}{3}+\frac{4}{3} \right) =4$$

216818_208290_ans.PNG

Mathematics

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