The area (in sq. units) of the region in the first quadrant bounded by the curve \(y^{2} = 4-x\) and the coordinate axes is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{16}{3}\)
The curve is a parabola opening to the left with vertex (4, 0). It meets the y-axis where x = 0:
\(y^{2} = 4,\ \ y = 2\) (first quadrant)
In the first quadrant the curve is \(y = \sqrt{4-x}\) for \(0 \le x \le 4\), so
\( \displaystyle A = \int_{0}^{4}\sqrt{4-x}\,dx \)
\( \displaystyle = \left[-\frac{2}{3}(4-x)^{3/2}\right]_{0}^{4} \)
\( \displaystyle = 0-\left(-\frac{2}{3}\times8\right) = \frac{16}{3} \)