The integral \(\displaystyle\int_{0}^{5}\frac{2}{5}\sqrt{25-x^{2}}\,dx\) gives the area of a region bounded by a curve and the coordinate axes. The value of the integral is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{5\pi}{2}\)
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:
\( \displaystyle I = \frac{2}{5}\left[\frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_{0}^{5} \)
\( \displaystyle = \frac{2}{5}\left(0+\frac{25}{2}\times\frac{\pi}{2}\right) \)
\( \displaystyle = \frac{2}{5}\times\frac{25\pi}{4} = \frac{5\pi}{2} \)
Shortcut: The curve \(\displaystyle y = \frac{2}{5}\sqrt{25-x^{2}}\) squares to \(\displaystyle\frac{x^{2}}{25}+\frac{y^{2}}{4} = 1\) with \(y \ge 0\). From x = 0 to x = 5 it bounds the quarter of the ellipse in the first quadrant, so the value is \(\displaystyle\frac{1}{4}\pi\times5\times2 = \frac{5\pi}{2}\).
Why the other options are incorrect
Option (1) \(5\pi\) — Incorrect. This is half of the ellipse; the limits 0 to 5 with \(y \ge 0\) give only a quarter.
Option (2) \(10\pi\) — Incorrect. This is the area of the whole ellipse.
Option (3) \(\displaystyle\frac{25\pi}{4}\) — Incorrect. This is \(\displaystyle\int_{0}^{5}\sqrt{25-x^{2}}\,dx\), the quarter circle; the factor \(\displaystyle\frac{2}{5}\) is left out.