Question

express the limit as a definite integral. \\(\lim_{n\to\infty} \frac{1}{n} \sum_{i=1}^{n} \frac{9}{1 + \left(\frac{i}{n}\
ight)^6}\\) \\(\int_{0}^{1} (\quad) dx\\)

Explanation:

Step1: Recall Riemann sum formula

For $\int_{a}^{b} f(x)dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x$, where $\Delta x = \frac{b-a}{n}$, $x_i = a + i\Delta x$

Step2: Identify $\Delta x$ and $x_i$

Here, $\Delta x = \frac{1}{n}$, $a=0$, $b=1$, so $x_i = 0 + i\cdot\frac{1}{n} = \frac{i}{n}$

Step3: Match to find $f(x)$

$f(x_i) = \frac{9}{1+(x_i)^6}$, so $f(x) = \frac{9}{1+x^6}$

Answer:

$\int_{0}^{1} \frac{9}{1+x^6} dx$