The area (in square units) enclosed between the curve \(x^2 = 4y\) and the line \(x = y\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{8}{3}\)
Putting y = x in \(x^2 = 4y\) gives the points of intersection:
\(x^2 = 4x\)
\(x(x - 4) = 0,\quad x = 0 \text{ or } 4\)
So the curves meet at (0, 0) and (4, 4). Between them the line \(y = x\) is above the parabola \(y = \frac{x^2}{4}\) (at x = 2: 2 > 1):
\( \displaystyle A = \int_{0}^{4}\left(x - \frac{x^2}{4}\right)dx \)
\( \displaystyle = \left[\frac{x^2}{2} - \frac{x^3}{12}\right]_{0}^{4} \)
\( \displaystyle = 8 - \frac{64}{12} = 8 - \frac{16}{3} \)
\( \displaystyle = \frac{8}{3} \)