Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 24

Evaluate $\displaystyle \int^5_0 \left( 1 + 2x^3 \right) dx$ using the form of the definition of the intergral $\displaystyle \int^b_a f(x) dx = \lim_{n \to \infty} \sum\limits_{i = 1}^n f(x_i) \Delta x$

$
\begin{equation}
\begin{aligned}
\Delta x &= \frac{b-a}{n} \\
\\
\Delta x &= \frac{5-0}{n}\\
\\
\Delta x &= \frac{5}{n}\\
\\
x_i &= a + i\Delta x\\
\\
x_i &= 0 + \frac{5i}{n}\\
\\
x_i &= \frac{5i}{n}
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n f \left( \frac{5i}{n} \right) \left( \frac{5}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ 1+2 \left(\frac{3i}{n}\right)^3 \right] \left( \frac{5}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ 1 + 2 \left( \frac{125i^3}{n^3}\right) \right] \left( \frac{5}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( 1 + \frac{250i^3}{n^3} \right) \frac{5}{n}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( \frac{5}{n} + \frac{1250 i^3}{n^4} \right)
\end{aligned}
\end{equation}
$


Evaluate the summation

$
\begin{equation}
\begin{aligned}
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{5}{n} + \sum\limits_{i = 1}^n \frac{1250 i^3}{n^4}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{1}{n} + \sum\limits_{i = 1}^n + \frac{1250}{n^4} \sum\limits_{i = 1}^n i^3\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{1}{\cancel{n}} \left(5 \cancel{n} \right) + \frac{1250}{n^4} \left[ \frac{n(n+1)}{2} \right]^2\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n 5 + \frac{1250}{n^4} \left[ \frac{\left( n^2 + n \right)^2}{4} \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n 5 + \frac{1250}{n^4} \left( \frac{n^4 + 2n^3 + n^2}{4} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n 5+ \frac{625\left(n^4 + 2n^3 + n^2 \right)}{2n^4}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n 5+ \frac{625n^2 \left(n^2+2n+1 \right)}{2n^4}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n 5+ \frac{625\left( n^2 + 2n +1 \right)}{2n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{10n^2 + 625n^2 + 1250n + 625}{2n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{635n^2 + 1250n + 625}{2n^2}
\end{aligned}
\end{equation}
$


Evaluating the limit

$
\begin{equation}
\begin{aligned}
\int^5_0 \left( 1 + 2x^3 \right) dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n f (x_i) \Delta x\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \lim_{n \to \infty} \left( \frac{635n^2 + 1250 n + 625}{2n^2} \right)\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \lim_{n \to \infty} \left( \frac{\frac{635\cancel{n^2}}{\cancel{n^2}}+\frac{1250n}{n^2}+\frac{625}{n^2}}{\frac{2\cancel{n^2}}{\cancel{n^2}}} \right)\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \lim_{n \to \infty} \left( \frac{635+\frac{1250}{n} + \frac{625}{n^2} }{2} \right)\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \frac{635 + \lim\limits_{n \to \infty}\frac{1250}{n}+\lim\limits_{n \to \infty} \frac{625}{n^2} }{2}\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \frac{635+0+0}{2}\\
\\
\int^5_0 \left( 1 + 2x^3 \right) dx &= \frac{635}{2} \qquad \text{ or } \qquad \int^5_0 \left( 1 + 2x^3 \right) dx = 317.5
\end{aligned}
\end{equation}
$