Area Under the Curve
Area of region with two parabolas
Grade 12
Question:
<p>The area (in sq. units) bounded by \(y^2=x\) and \(x^2=y\). [JEE Main 2019]</p>
<li>\(\dfrac{1}{3}\)</li>
<li>\(\dfrac{2}{3}\)</li>
<li>\(\dfrac{1}{6}\)</li>
<li>\(\dfrac{3}{2}\)</li>
Step-by-Step Solution
Key Concept: The two parabolas y^2=x and x^2=y meet at (0,0) and (1,1). Area = \int_0^1(\sqrt{x} - x^2)dx = 2/3 - 1/3 = 1/3.
<div class='solution'>
<p>Intersections: $y^2=x$ and $y=x^2\Rightarrow(x^2)^2=x\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0,1$.</p>
<p>On $[0,1]$: upper curve is $y=\sqrt{x}$, lower is $y=x^2$.</p>
<p>$$A=\int_0^1(\sqrt{x}-x^2)dx=\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_0^1=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$$</p>
</div>
Correct Answer: A