Question

Let \(\alpha>0\) If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :

• A. 2
• B. $2 \sqrt{2}$
• C. 4
• D. $\sqrt{2}$

Hint:

When solving problems with integrals involving square roots, consider rationalizing the expression to simplify the calculation and isolate the variable you're solving for.

Answer 1

The Correct Option is A

After rationalising
\(\int\limits_{0}^{a}\frac{x}{\alpha}\left(\sqrt{x+\alpha}+\sqrt{x}\right)dx\)
\(=\int\limits_{0}^{a}\frac{1}{\alpha}[(x+\alpha)^{3/2}-\alpha(x+\alpha)^{1/2}+x^{3/2}]dx\)
\(=\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5/2}-\alpha\frac{2}{3}(x+\alpha)^{3/2}+\frac{2}{5}x^{5/2}\right]|^a_{0}\)
\(=\frac{1}\alpha{}[\frac{2}{5}(2\alpha)^{5/2}-\frac{2\alpha}{3}(2\alpha)^{3/2}+\frac{2}{5}\alpha^{5/2}-\frac{2}{5}\alpha^{5/2}+\frac{2}{3}\alpha^{5/2}]|^{a}_0\)
\(=\frac{1}{\alpha}[\frac{2^{7/2}\alpha^{5/2}}{5}-\frac{2^{5/2}\alpha^{5/2}}{3}+\frac{2}{3}\alpha^{5/2}]\)
\(=\alpha^{3/2}[\frac{2^{7/2}}{5}-\frac{2^{5/2}}{3}+\frac{2}{3}]\)
\(=\frac{\alpha^{3/2}}{15}(24\sqrt2-20\sqrt2+10)\)
\(=\frac{\alpha^{3/2}}{15}(4\sqrt2+10)\)
Now,
\(\frac{\alpha^{3/2}}{15}(4\sqrt2+10)=\frac{16+20\sqrt2}{15}\)
\(⇒\alpha=2\)
So, the correct option is (A) : 2

Answer 2

We are given the integral: \[ \int_{\alpha}^{x} \left( \sqrt{x + \alpha + \sqrt{x}} \right) \, dx = \frac{16 + 20 \sqrt{2}}{15}. \] Step 1: After rationalizing the integral, we obtain: \[ \int_{\alpha}^{x} \left[ (x + \alpha)^2 - \alpha(x + \alpha)^2 + x^2 \right] \, dx. \] This simplifies to: \[ \frac{1}{\alpha} \left[ \frac{2x^2}{5} - \frac{2}{3} (x + \alpha)^2 + \frac{2}{5} x^2 + 2 \right]. \] Step 2: Simplifying further: \[ \frac{1}{\alpha} \left[ \frac{5}{2} (2x^2) - \frac{2}{3} (x + \alpha)^2 + \frac{2}{5} x^2 + 2 \right]. \] Step 3: Now, we get: \[ \frac{1}{\alpha} \left[ \frac{5}{2} (x) - \frac{2}{5} (x + \alpha) \right] \quad \text{this simplifies further to } \alpha = 2. \]