1. We need to evaluate $I=\int_{\frac{3\sqrt2}{4}}^{\frac{3\sqrt3}{4}} \frac{48}{\sqrt{9-4x^2}}\,dx.$

2. Use the standard form $\int \frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)\cdot \frac{1}{u'} \quad \text{(with suitable substitution)}.$

Here, $9-4x^2=3^2-(2x)^2.$ So let $u=2x \implies du=2dx \implies dx=\frac{du}{2}.$

Then $I=\int \frac{48}{\sqrt{9-u^2}}\cdot \frac{du}{2}=24\int \frac{du}{\sqrt{9-u^2}}.$

3. Now, $\int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right).$ Hence, $I=24\sin^{-1}\left(\frac{u}{3}\right)\Bigg|_{u=2x} = 24\sin^{-1}\left(\frac{2x}{3}\right)\Bigg|_{\frac{3\sqrt2}{4}}^{\frac{3\sqrt3}{4}}.$

4. Substitute the limits.

At $x=\frac{3\sqrt3}{4},$ we get $\frac{2x}{3}=\frac{2}{3}\cdot \frac{3\sqrt3}{4}=\frac{\sqrt3}{2},$ so $\sin^{-1}\left(\frac{\sqrt3}{2}\right)=\frac{\pi}{3}.$

At $x=\frac{3\sqrt2}{4},$ we get $\frac{2x}{3}=\frac{2}{3}\cdot \frac{3\sqrt2}{4}=\frac{\sqrt2}{2},$ so $\sin^{-1}\left(\frac{\sqrt2}{2}\right)=\frac{\pi}{4}.$

Therefore, $I=24\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=24\left(\frac{4\pi-3\pi}{12}\right)=24\cdot \frac{\pi}{12}=2\pi.$

5. Hence the value of the integral is $\boxed{2\pi}.$

6. Option check:

• A: $\frac{\pi}{2}$ ❌
• B: $\frac{\pi}{3}$ ❌
• C: $\frac{\pi}{6}$ ❌
• D: $2\pi$ ✅

7. Comparison with stored answer: Stored correct answer is D, which matches our result.