The area (in sq. units) of the region in the first quadrant bounded b…

The correct answer is Option (2) → \(5\pi\)

In the first quadrant, the upper half of the ellipse gives

    \( \displaystyle y = \frac{4}{5}\sqrt{25-x^{2}},\quad 0 \le x \le 5 \)

So the area is

    \( \displaystyle A = \frac{4}{5}\int_{0}^{5}\sqrt{25-x^{2}}\,dx \)

Using \( \displaystyle \int\sqrt{a^{2}-x^{2}}\,dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}+C \) with a = 5:

    \( \displaystyle A = \frac{4}{5}\left[\frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_{0}^{5} \)

    \( \displaystyle = \frac{4}{5}\left(0+\frac{25}{2}\times\frac{\pi}{2}\right) \)

    \( \displaystyle = \frac{4}{5}\times\frac{25\pi}{4} = 5\pi \)

Shortcut: The whole ellipse has area \(\pi ab = \pi \times 5 \times 4 = 20\pi\), and the region is one quarter of it, so the area is \(5\pi\).