The area (in sq. units) of region bounded by the curves \( y^2=4x \) and \( y=x \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8/3
Putting y = x in \( y^2=4x \) gives \( x^2=4x \), so the curves meet at x = 0 and x = 4. Between them the parabola \( y=2\sqrt{x} \) lies above the line y = x.
\( \displaystyle \text{Area}=\int_0^4(2\sqrt{x}-x)\,dx \)
\( \displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{x^2}{2}\right]_0^4 \)
\( \displaystyle =\frac{32}{3}-8 \)
\( \displaystyle =\frac{8}{3} \)