Question

The value of the integral:$\int\limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x$ is equal to

• A. $2 \pi$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

The Correct Option is A

Step 1: Simplify the integrand 

The given integral is:

\[ I = \int_{ \frac{3 \sqrt{3}}{4} }^{ \frac{3 \sqrt{2}}{4} } 48 \sqrt{9 - 4x^2} \, dx. \]

The term \( \sqrt{9 - 4x^2} \) suggests the substitution:

\[ x = \frac{3}{2} \sin \theta \quad \text{so} \quad dx = \frac{3}{2} \cos \theta \, d\theta. \]

Under this substitution:

\[ 9 - 4x^2 = 9 - 4 \left( \frac{3}{2} \sin \theta \right)^2 = 9 - 9 \sin^2 \theta = 9 \cos^2 \theta. \] Thus: \[ \sqrt{9 - 4x^2} = 3 \cos \theta. \] 

Step 2: Transform the limits

We now transform the limits of integration:

• When \( x = \frac{3 \sqrt{3}}{4} \), we have: \[ \frac{3}{2} \sin \theta = \frac{3 \sqrt{3}}{4} \quad \Rightarrow \quad \sin \theta = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \theta = \frac{\pi}{3}. \]
• When \( x = \frac{3 \sqrt{2}}{4} \), we have: \[ \frac{3}{2} \sin \theta = \frac{3 \sqrt{2}}{4} \quad \Rightarrow \quad \sin \theta = \frac{\sqrt{2}}{2} \quad \Rightarrow \quad \theta = \frac{\pi}{4}. \]

Step 3: Substitute into the integral The integral becomes: \[ I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} 48 \cdot 3 \cos \theta \cdot \frac{3}{2} \cos \theta \, d\theta. \] Simplifying: \[ I = \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} 24 \, d\theta. \] 

Step 4: Evaluate the integral \[ I = 24 \left[ \theta \right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} = 24 \left( \frac{\pi}{4} - \frac{\pi}{3} \right). \] Simplifying the terms: \[ \frac{\pi}{4} - \frac{\pi}{3} = \frac{3\pi - 4\pi}{12} = -\frac{\pi}{12}. \] Thus: \[ I = 24 \times \left( -\frac{\pi}{12} \right) = -2\pi. \] 

Final Answer: \[ I = 2\pi. \]