Question

Evaluate: \(\displaystyle \int_0^a \left( \sqrt{x} + \sqrt{a - x} \right) \sqrt{x} \, dx = \)

• A. \(a\)
• B. \(2a\)
• C. \( \frac{a}{2} \)
• D. \(3a\)

Hint:

For integrals involving square roots of linear expressions, substitution and Beta/Gamma functions can help evaluate definite integrals.

Answer 1

The Correct Option is B

Step 1: Rewrite the integrand: \[ \left(\sqrt{x} + \sqrt{a - x}\right) \sqrt{x} = x + \sqrt{x(a - x)} \] Step 2: Split the integral: \[ \int_0^a \left( x + \sqrt{x(a - x)} \right) dx = \int_0^a x \, dx + \int_0^a \sqrt{x(a - x)} \, dx \] Step 3: Evaluate the first integral: \[ \int_0^a x \, dx = \frac{a^2}{2} \] Step 4: For the second integral, substitute \(x = a t\), so \(dx = a dt\), limits change from \(x=0 \to t=0\) and \(x=a \to t=1\): \[ \int_0^a \sqrt{x(a - x)} \, dx = \int_0^1 \sqrt{a t (a - a t)} a dt = a^{2} \int_0^1 \sqrt{t(1 - t)} dt \] Step 5: The integral \(\int_0^1 \sqrt{t(1-t)} dt\) is a Beta function \(B(\frac{3}{2}, \frac{3}{2})\) and equals \[ B\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\Gamma(\frac{3}{2})^2}{\Gamma(3)} = \frac{\left(\frac{\sqrt{\pi}}{2}\right)^2}{2} = \frac{\pi}{8} \] Step 6: So, \[ \int_0^a \sqrt{x(a - x)} \, dx = a^{2} \cdot \frac{\pi}{8} \] Step 7: Therefore, \[ \int_0^a \left( \sqrt{x} + \sqrt{a - x} \right) \sqrt{x} \, dx = \frac{a^2}{2} + \frac{\pi a^{2}}{8} \] However, none of the options directly match this. --- Check the problem statement or options again? If instead the integral is: \[ \int_0^a \left( \sqrt{x} + \sqrt{a - x} \right) dx \] or something similar, let me know! ---