Definite Integration
Grade 12
Question:
<p>The value of <span class="math-tex">\(\lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum \limits_{r=0}^{2 n-1} \frac{n^2}{n^2+4 r^2}\)</span> is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2} \tan ^{-1}(4)\)</span></p>
<p style="display:inline">tan<sup>-1</sup>(4)</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{4} \tan ^{-1}(4)\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2} \tan ^{-1}(2)\)</span></p>
Step-by-Step Solution
Key Concept: The limit of a Riemann sum can be evaluated by converting it into a definite integral where r/n is replaced by x, 1/n by dx, and the limits are determined by the range of r/n.
<p><span class="math-tex">$L=\lim \limits_{n \rightarrow \infty} \cdot \frac{1}{n} \cdot \sum \limits_{r=0}^{2 n-1} \frac{1}{1+4\left(\frac{r}{n}\right)^2}$</span><br />
<span class="math-tex">$\Rightarrow \mathrm{L}=\int \limits_0^2 \frac{1}{1+4 \mathrm{x}^2} \mathrm{dx} $</span> <span class="math-tex">$\Rightarrow \frac{1}{4} \int \limits_0^2 \frac{\mathrm{dx}}{\left(\frac{1}{2}\right)^2+\mathrm{x}^2}$</span><br />
<span class="math-tex">$\Rightarrow L=\left[\frac{1}{2} \tan ^{-1}(2 x)\right]_0^2=\frac{1}{2} \tan ^{-1} 4$</span></p>
Correct Answer: A