Question

The value of \[ \int_{\alpha}^{\beta} \sqrt{(x-\alpha)(\beta-x)}\,dx, \qquad \alpha\ne\beta \] is

• A. \((\beta-\alpha)\)
• B. \((\beta-\alpha)^2\)
• C. \[ \frac{\pi}{2}(\beta-\alpha)^2 \]
• D. \[ \frac{\pi}{8}(\beta-\alpha)^2 \]

Hint:

A standard result: \[ \int_a^b \sqrt{(x-a)(b-x)}\,dx = \frac{\pi}{8}(b-a)^2. \] This formula is frequently asked in entrance examinations.

Answer 1

The Correct Option is D

Concept: Use the substitution \[ x=\frac{\alpha+\beta}{2} +\frac{\beta-\alpha}{2}\sin\theta. \] This transforms the integral into a standard trigonometric integral.

Step 1: Apply the substitution. Let \[ a=\frac{\beta-\alpha}{2}. \] Then \[ x=\frac{\alpha+\beta}{2}+a\sin\theta \] and \[ dx=a\cos\theta\,d\theta. \] Also, \[ x-\alpha=a(1+\sin\theta), \] \[ \beta-x=a(1-\sin\theta). \] Hence \[\begin{aligned} \sqrt{(x-\alpha)(\beta-x)} &= a\cos\theta. \end{aligned}\]

Step 2: Transform the integral. When \[ x=\alpha, \] \[ \theta=-\frac{\pi}{2} \] and when \[ x=\beta, \] \[ \theta=\frac{\pi}{2}. \] Therefore, \[\begin{aligned} I &= a^2 \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta. \end{aligned}\]

Step 3: Evaluate the integral. \[\begin{aligned} \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta = \frac{\pi}{2} \end{aligned}\] Hence, \[\begin{aligned} I &= a^2\cdot\frac{\pi}{2} \\ &= \frac{\pi}{2} \left( \frac{\beta-\alpha}{2} \right)^2 \\ &= \frac{\pi}{8} (\beta-\alpha)^2. \end{aligned}\] \[\begin{aligned} \boxed{ \frac{\pi}{8}(\beta-\alpha)^2 } \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.