Question

\( \int_{0}^{2} \sqrt{(x+3)(2-x)} d x= \)

• A. \( \frac{25}{8} \cos ^{-1}\left(\frac{1}{5}\right)-\frac{\sqrt{6}}{4} \)
• B. \( \frac{25}{8} \sin ^{-1}\left(\frac{1}{5}\right)-\frac{\sqrt{6}}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \pi \)

Hint:

Completing the square allows the use of standard integral forms. Be mindful of identities to convert answers to the form given in options.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

We evaluate the definite integral by completing the square for the quadratic inside the root, converting it to \( \sqrt{a^2 - X^2} \), and applying standard integration formulas.
Step 2: Key Formula or Approach:

1. \( \int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(x/a) \). 2. \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).
Step 3: Detailed Explanation:

Expand integrand: \( (x+3)(2-x) = 2x - x^2 + 6 - 3x = -x^2 - x + 6 \). Complete the square: \( -(x^2 + x - 6) = -\left( (x+1/2)^2 - 1/4 - 6 \right) = \frac{25}{4} - (x+1/2)^2 \). Let \( t = x + 1/2 \). \( dt = dx \). Limits: \( x=0 \to t=1/2 \); \( x=2 \to t=5/2 \). \( I = \int_{1/2}^{5/2} \sqrt{(5/2)^2 - t^2} dt \). Formula gives: \( \left[ \frac{t}{2}\sqrt{\frac{25}{4}-t^2} + \frac{25}{8}\sin^{-1}\left(\frac{t}{5/2}\right) \right]_{1/2}^{5/2} \). Upper limit (\( t=5/2 \)): Term 1: \( \frac{5}{4}\sqrt{0} = 0 \). Term 2: \( \frac{25}{8}\sin^{-1}(1) = \frac{25\pi}{16} \). Lower limit (\( t=1/2 \)): Term 1: \( \frac{1}{4}\sqrt{\frac{25}{4}-\frac{1}{4}} = \frac{1}{4}\sqrt{6} = \frac{\sqrt{6}}{4} \). Term 2: \( \frac{25}{8}\sin^{-1}\left(\frac{1/2}{5/2}\right) = \frac{25}{8}\sin^{-1}(1/5) \). So, \( I = \frac{25\pi}{16} - \left( \frac{\sqrt{6}}{4} + \frac{25}{8}\sin^{-1}(1/5) \right) \). \( I = \frac{25}{8} \left( \frac{\pi}{2} - \sin^{-1}(1/5) \right) - \frac{\sqrt{6}}{4} \). Using identity: \( I = \frac{25}{8} \cos^{-1}(1/5) - \frac{\sqrt{6}}{4} \).
Step 4: Final Answer:

Matches Option (A).