Area Under the Curve
Area with inverse function (advanced)
Grade 12
Question:
<p>The area bounded by \(y=f(x)\), \(y=f^{-1}(x)\) and \(y=x\). Assume \(f:[0,1]\to[0,1]\) with \(f(x)=x^2\). [JEE Advanced 2005]</p>
Step-by-Step Solution
Key Concept: y=f(x) and y=f⁻^1(x) are reflections about y=x. Both regions between f(x) and y=x (and f⁻^1(x) and y=x) are equal. Total = 2 \cdot \int_0^1(x-x^2)dx = 2 \cdot (1/6) = 1/3.
<div class='solution'>
<p>$f(x)=x^2$, $f^{-1}(x)=\sqrt{x}$ on $[0,1]$.</p>
<p>Area between $y=\sqrt{x}$ and $y=x$: $\int_0^1(\sqrt{x}-x)dx=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$.</p>
<p>Area between $y=x$ and $y=x^2$: $\int_0^1(x-x^2)dx=\frac{1}{6}$.</p>
<p>Total area enclosed by all three curves: $\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$. ✓</p>
</div>
Correct Answer: A