The area (in sq. units) of the region enclosed by the parabola \(y = x^{2}\) and the line y = 4x is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{32}{3}\)
The points of intersection come from \(x^{2} = 4x\), that is x = 0 and x = 4. Between them the line is above the parabola (at x = 1 the line gives 4 and the parabola gives 1), so the area is the integral of upper minus lower:
\( \displaystyle A = \int_{0}^{4}(4x-x^{2})\,dx \)
\( \displaystyle = \left[2x^{2}-\frac{x^{3}}{3}\right]_{0}^{4} \)
\( \displaystyle = 32-\frac{64}{3} = \frac{32}{3} \)