Definite Integration
Grade 12
Question:
<p>Let <span class="math-tex">\(\alpha\)</span> > - 1 and <span class="math-tex">\(\beta\)</span> > - 1, then the value of <span class="math-tex">\(\lim _\limits{n \rightarrow \infty} n^{\beta-\alpha}\left(\frac{1^{\alpha}+2^{\alpha}+\ldots+n^{\alpha}}{1^{\beta}+2^{\beta}+\ldots+n^{\beta}}\right)\)</span> is :</p>
<p style="display:inline"><span class="math-tex">\(\frac{\beta+1}{\alpha+1}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\beta+2}{\alpha+2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\alpha+2}{\beta+2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\alpha+1}{\beta+1}\)</span></p>
Step-by-Step Solution
<p><span class="math-tex">$\mathop {\lim }\limits_{n \to \infty } \frac{{{n^\beta }}}{{{n^\alpha }}}\frac{{\sum\limits_{r = 1}^n {{r^\alpha }} }}{{\sum\limits_{r = 1}^n {{r^\beta }} }} = \frac{{\int_0^1 {{x^\alpha }} dx}}{{\int_0^1 {{x^\beta }} dx}} = \frac{{\beta + 1}}{{\alpha + 1}}$</span></p>
Correct Answer: A