Definite Integration
Grade None
Question:
<p>If I <span class="math-tex">\((m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x, m, n \gt 0\)</span>, then <span class="math-tex">\(I(9,14)+I(10,13)\)</span> is</p>
<p style="display:inline"><span class="math-tex">\({I}(1,13)\)</span></p>
<p style="display:inline"><span class="math-tex">\({I}(9,13)\)</span></p>
<p style="display:inline"><span class="math-tex">\({I}(19,27)\)</span></p>
<p style="display:inline"><span class="math-tex">\({I}(9,1)\)</span></p>
Step-by-Step Solution
Key Concept: Convert the Beta functions into their trigonometric forms to factor out common terms and simplify the sum using the identity sin²θ + cos²θ = 1.
<p>Given <span class="math-tex">$I(m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$</span><br />
Put <span class="math-tex">$x=\sin ^{2} \theta \Rightarrow d x=2 \sin \theta \cos \theta d \theta$</span><br />
When <span class="math-tex">$x=0$</span> then <span class="math-tex">$\theta=0$</span><br />
When <span class="math-tex">$x=1$</span> then <span class="math-tex">$\theta=\frac{\pi}{2}$</span><br />
Now<br />
<span class="math-tex">$I(m, n)=2 \int_{0}^{\pi / 2}(\sin \theta)^{2 m-1}(\cos \theta)^{2 n-1} d \theta$</span><br />
<span class="math-tex">$I(9,14)+I(10,13)$</span><br />
<span class="math-tex">$=2 \int_{0}^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{27} d \theta$</span><span class="math-tex">$+2 \int_{0}^{\pi / 2}(\sin \theta)^{19}(\cos \theta)^{25} {~d} \theta$</span><br />
<span class="math-tex">$=2 \int_{0}^{\pi / 2}(\sin \theta)^{17}(\cos \theta)^{25} d \theta=I(9,13)$</span></p>
Correct Answer: B