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$O A B C$ is a unit square where $O$ is the origin and $B=(1,1)$. The curves $y^2=x$ and $x^2=y$ divide the area of the square into three parts $O A B O, O B O$ and $O B C O$. If $a_1, a_2, a_3$ are the areas (in sq units) of these parts respectively, then $a_1+2 a_2+3 a_3=$
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The correct answer is:
2
Now, $\begin{aligned} a_2 & =\int_0^1\left(\sqrt{x}-x^2\right) d x \\ & =\left[\frac{2}{3} x^{3 / 2}-\frac{x^3}{3}\right]_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\end{aligned}$
So, $\quad a_1+a_3+\frac{1}{3}=1$
From Eqs. (ii) and (iii),
$$
a_1=a_3=\frac{1}{3}=a_2
$$
So, $a_1+2 a_2+3 a_3=\frac{1}{3}+\frac{2}{3}+\frac{3}{3}=2$
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