Definite Integration
Grade 12
Question:
<p><span class="math-tex">\(\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x\)</span> is equal to</p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{6}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2\pi\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{3}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\pi}{2}\)</span></p>
Step-by-Step Solution
Key Concept: Recognize the integrand as a standard form ∫dx/√(a²-x²) = sin⁻¹(x/a) + C by rewriting 9-4x² as 9-(2x)², then apply the antiderivative formula with appropriate substitution u=2x to evaluate the definite integral.
<p><span class="math-tex">\(\int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x\)</span> we know <span class="math-tex">\( \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+c\)</span><br />
So, <span class="math-tex">\( \int \limits_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} \mathrm{dx}=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 \mathrm{x}}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}\)</span><br />
<span class="math-tex">\(=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]\)</span><br />
<span class="math-tex">\(=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]\)</span><br />
<span class="math-tex">\(=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=24 \times \frac{\pi}{12}=2 \pi\)</span></p>
Correct Answer: B