Definite Integration as a Limit of a Sum
General
Grade 12
Question:
Evaluate $\lim_{n \to \infty} \left[ \frac{\sqrt{n}}{(3+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2}+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{3}(3\sqrt{3}+4\sqrt{n})^2} + \dots + \frac{1}{49n} \right]$
Step-by-Step Solution
Key Concept: General
Let $p = \lim_{n \to \infty} \left[ \frac{\sqrt{n}}{(3+4\sqrt{n})^2} + \frac{\sqrt{n}}{\sqrt{2}(3\sqrt{2}+4\sqrt{n})^2} + \dots + \frac{\sqrt{n}}{\sqrt{n}(3\sqrt{n}+4\sqrt{n})^2} \right]$<br>Analyzing the expression with the view of increasing integral value we get the expression in terms of $r$ as<br>$= \lim_{n \to \infty} \sum_{r=1}^n \frac{\sqrt{n}}{\sqrt{r}(3\sqrt{r}+4\sqrt{n})^2} = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n \sqrt{\frac{r}{n}} \left( 3\sqrt{\frac{r}{n}} + 4 \right)^2} = \int_0^1 \frac{dx}{\sqrt{x}(3\sqrt{x}+4)^2}$<br>Put $3\sqrt{x}+4=t, \therefore \frac{3}{2\sqrt{x}}dx = dt$<br>Hence $p = \frac{2}{3} \int_4^7 \frac{dt}{t^2} = \frac{2}{3} \left[ -\frac{1}{t} \right]_4^7 = \frac{2}{3} \left( -\frac{1}{7} + \frac{1}{4} \right) = \frac{1}{14}$
Correct Answer: 1/14