Question

If $$\int_{0}^{a} \sqrt{\frac{a-x}{x}} \, dx = \frac{k}{2}$$ then $k =$

• A. $\pi a$
• B. $\pi a^2$
• C. $\frac{5\pi a}{2}$
• D. $\frac{3\pi a}{2}$

Hint:

You can use Walli's formula shortcut for the final step: $\int_{0}^{\frac{\pi}{2}} \cos^2\theta \, d\theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}$. Multiplying this by $2a$ gives $2a\left(\frac{\pi}{4}\right) = \frac{\pi a}{2}$ in one step, bypassing full integration!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a definite integral involving a square root fraction bounded between $0$ and $a$. The value of this integral is equated to $\frac{k}{2}$. We need to evaluate the integral to solve for the parameter $k$.

Step 2: Key Formula or Approach:
For integrals containing the algebraic group structure $\sqrt{\frac{a-x}{x}}$, the standard trigonometric substitution is: $$x = a\sin^2\theta \implies dx = 2a\sin\theta\cos\theta \, d\theta$$ This substitution simplifies the expression under the radical using the identity $1 - \sin^2\theta = \cos^2\theta$.

Step 3: Detailed Explanation:
Let's apply the substitution $x = a\sin^2\theta$ to our integral: $$dx = 2a\sin\theta\cos\theta \, d\theta$$ Now let's change our boundary limits of integration: 1. Lower limit: When $x = 0 \implies a\sin^2\theta = 0 \implies \theta = 0$. 2. Upper limit: When $x = a \implies a\sin^2\theta = a \implies \sin^2\theta = 1 \implies \theta = \frac{\pi}{2}$. Substitute these values into the definite integral: $$I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a - a\sin^2\theta}{a\sin^2\theta}} \cdot (2a\sin\theta\cos\theta) \, d\theta$$ Factor out $a$ inside the square root and simplify: $$I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{a(1 - \sin^2\theta)}{a\sin^2\theta}} \cdot (2a\sin\theta\cos\theta) \, d\theta$$ $$I = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{\cos^2\theta}{\sin^2\theta}} \cdot (2a\sin\theta\cos\theta) \, d\theta$$ $$I = \int_{0}^{\frac{\pi}{2}} \left(\frac{\cos\theta}{\sin\theta}\right) \cdot (2a\sin\theta\cos\theta) \, d\theta$$ The $\sin\theta$ terms cancel out perfectly: $$I = 2a \int_{0}^{\frac{\pi}{2}} \cos^2\theta \, d\theta$$ Using the trigonometric identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$: $$I = 2a \int_{0}^{\frac{\pi}{2}} \left(\frac{1 + \cos 2\theta}{2}\right) d\theta = a \int_{0}^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta$$ Integrate each term: $$I = a \left[ \theta + \frac{\sin 2\theta}{2} \right]_{0}^{\frac{\pi}{2}}$$ Evaluate at the limits: $$I = a \left[ \left(\frac{\pi}{2} + \frac{\sin \pi}{2}\right) - (0 + 0) \right] = a \left( \frac{\pi}{2} + 0 \right) = \frac{\pi a}{2}$$ We are given that the integral value is equal to $\frac{k}{2}$: $$\frac{\pi a}{2} = \frac{k}{2} \implies k = \pi a$$

Step 4: Final Answer:
The value of $k$ is $\pi a$, which corresponds to option (A).