Question

Evaluate: \(\int_0^3 \sqrt{9 - x^2} \, dx\).

Hint:

Geometry shortcut: \(\int_0^r \sqrt{r^2 - x^2} \, dx\) is simply the area of a quadrant of a circle, which is \(\frac{1}{4} \pi r^2\). Here \(r=3\), so Area \(= \frac{1}{4} \pi (3)^2 = \frac{9\pi}{4}\). This is much faster than integration!

Answer 1

Step 1: Understanding the Concept:
The integral represents the area of a quarter-circle of radius 3 in the first quadrant. It can be solved using the standard formula for \(\int \sqrt{a^2 - x^2} \, dx\).

Step 2: Key Formula or Approach:
Formula: \(\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C\).

Step 3: Detailed Explanation:
Here \(a = 3\).
\[ \int_0^3 \sqrt{9 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) \right]_0^3 \] Substitute the upper limit \(x = 3\):
\[ \left( \frac{3}{2}\sqrt{9 - 9} + \frac{9}{2}\sin^{-1}(1) \right) = 0 + \frac{9}{2} \left( \frac{\pi}{2} \right) = \frac{9\pi}{4} \] Substitute the lower limit \(x = 0\):
\[ \left( \frac{0}{2}\sqrt{9 - 0} + \frac{9}{2}\sin^{-1}(0) \right) = 0 + 0 = 0 \] Final result: \(\frac{9\pi}{4} - 0 = \frac{9\pi}{4}\).

Step 4: Final Answer:
The value of the integral is \(\frac{9\pi}{4}\).