Definite Integration
Grade None
Question:
<p>The value of <span class="math-tex">\(k \in N\)</span> for which the integral <span class="math-tex">\(I_{n}=\int_{0}^{1}\left(1-x^{k}\right)^{n} d x, n \in \mathbb{N}\)</span>, satisfies <span class="math-tex">\(147 I_{20}=148 I_{21}\)</span> is:</p>
<p style="display:inline">10</p>
<p style="display:inline">7</p>
<p style="display:inline">8</p>
<p style="display:inline">14</p>
Step-by-Step Solution
Key Concept: Establish a recurrence relation for the integral by using integration by parts and rewriting the resulting integrand to relate $I_n$ to $I_{n-1}$.
<p>Given,<br />
<span class="math-tex">$\Rightarrow I_{n}=\int_{0}^{1}\left(1-x^{k}\right)^{n} \cdot 1 d x$</span><br />
<span class="math-tex">$\Rightarrow I_{n}=$</span><span class="math-tex">$\left(1-x^{k}\right)^{n} \cdot x-n k \int_{0}^{1}\left(1-x^{k}\right)^{n-1} \cdot x^{k-1} \cdot d x$</span><br />
<span class="math-tex">$\Rightarrow I_{n}=$</span><span class="math-tex">$n k \int_{0}^{1}\left[\left(1-x^{k}\right)^{n}-\left(1-x^{k}\right)^{n-1}\right] d x$</span><br />
<span class="math-tex">$\Rightarrow I_{n}=n k I_{n-1}-n k I_{n}$</span><br />
<span class="math-tex">$\Rightarrow \frac{I_{n}}{I_{n-1}}=\frac{n k}{n k+1}$</span><br />
<span class="math-tex">$\Rightarrow \frac{I_{21}}{I_{20}}=\frac{21 k}{1+21 k}=\frac{147}{148} \Rightarrow k=7$</span></p>
Correct Answer: B