Definite Integration
Grade None
Question:
<p>Let <span class="math-tex">\(\beta(m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x, m, n \gt 0\)</span>. If <span class="math-tex">\(\int_{0}^{1}\left(1-x^{10}\right)^{20} d x=a \times \beta(b, c)\)</span> then <span class="math-tex">\(100(a+b+x)\)</span> equals ________.</p>
<p style="display:inline">1021</p>
<p style="display:inline">2120</p>
<p style="display:inline">1120</p>
<p style="display:inline">2012</p>
Step-by-Step Solution
Key Concept: Transform the given integral into the standard Beta function form by applying the substitution $x^{10} = t$ and comparing the resulting exponents to the definition $\beta(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1} dx$.
<p>We know<br />
<span class="math-tex">$\beta(m, n)=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$</span><br />
Let <span class="math-tex">$I=\int_{0}^{1} 1 .\left(1-x^{10}\right)^{20} d x$</span><br />
Put <span class="math-tex">$x^{10}=t \Rightarrow x=t^{1 / 10}$</span><br />
<span class="math-tex">$\therefore d x=\frac{1}{10}(t)^{-9 / 10} d t$</span><br />
<span class="math-tex">$I=\int_{0}^{1}(1-t)^{20} \frac{1}{10}(t)^{-9 / 10} d t$</span><br />
<span class="math-tex">$I=\frac{1}{10} \int_{0}^{1} t^{-9 / 10}(1-t)^{20} d t$</span><br />
<span class="math-tex">$a=\frac{1}{10}, b=\frac{1}{10}, c=21$</span></p>
Correct Answer: B