Answer
$\lim\limits_{n \to \infty}\sum_{i=1}^{n}(2+\frac{8~i}{n})^6\cdot \frac{8}{n} = 1,428,553.14$
Work Step by Step
We can use the definition of the integral in Theorem 4 to evaluate the integral:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{10-2}{n} = \frac{8}{n}$
$x_i = 2+\frac{8~i}{n}$
$\int_{2}^{10}x^6~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}(2+\frac{8~i}{n})^6\cdot \frac{8}{n} = 1,428,553.14$