Answer
(a) $\lim\limits_{n \to \infty}\sum_{i=1}^{n}[(1+\frac{4i}{n})+2(1+\frac{4i}{n})^5]\cdot \frac{4}{n} = 5220$
(b) $\int_{1}^{5}(x+2x^5)~dx = 5220$
Work Step by Step
(a) We can express the integral as the limit of a Riemann sum:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$
$\Delta x = \frac{b-a}{n} = \frac{5-1}{n} = \frac{4}{n}$
$x_i^* = a+i~\Delta x = 1+\frac{4i}{n}$
Note that $x_i^*$ is the right endpoint of each subinterval.
$\int_{1}^{5}(x+2x^5)~dx$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}[(1+\frac{4i}{n})+2(1+\frac{4i}{n})^5]\cdot \frac{4}{n}$
$= 5220$
(b) We can use the Fundamental Theorem to evaluate the integral:
$\int_{1}^{5}(x+2x^5)~dx$
$= (\frac{x^2}{2}+\frac{x^6}{3})~\vert_{1}^{5}$
$= (\frac{5^2}{2}+\frac{5^6}{3})-(\frac{1^2}{2}+\frac{1^6}{3})$
$= (\frac{25}{2} +\frac{15,625}{3})-(\frac{1}{2}+\frac{1}{3})$
$= (\frac{75}{6} +\frac{31,250}{6})-(\frac{5}{6})$
$= \frac{31,320}{6}$
$= 5220$
