Area Under the Curve
Area = 12 (integer type)
Grade 12
Question:
<p>The area (in sq. units) bounded by \(y^2=4x\), y-axis and \(y=3\). [JEE Advanced 2013]</p>
<li>\(\dfrac{9}{4}\)</li>
<li>\(\dfrac{9}{2}\)</li>
<li>\(\dfrac{3}{4}\)</li>
<li>\(3\)</li>
Step-by-Step Solution
Key Concept: Area = \int_0^3 (y^2/4) dy = [y^3/12]_0^3 = 27/12 = 9/4.
<div class='solution'>
<p>The parabola $y^2=4x\Rightarrow x=y^2/4$. Area between x=0 (y-axis) and $x=y^2/4$ from $y=0$ to $y=3$:</p>
<p>$$A=\int_0^3\frac{y^2}{4}\,dy=\frac{1}{4}\cdot\frac{27}{3}=\frac{9}{4}$$</p>
</div>
Correct Answer: A