Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 22

Evaluate $\displaystyle \int^4_1 \left( x^2 + 2x - 5 \right) dx$ 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{4-1}{n} \\
\\
\Delta x &= \frac{3}{n}\\
\\
x_i &= a+ i \Delta x\\
\\
x_i &= 1 + \frac{3i}{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( 1 + \frac{3i}{n} \right) \left( \frac{3}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ \left( 1 + \frac{3i}{n}\right)^2 + 2\left( 1 + \frac{3i}{n}\right) - 5 \right] \left( \frac{3}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( 1 + \frac{6i}{n} + \frac{9i^2}{n^2} + 2 + \frac{6i}{n} - 5 \right) \left( \frac{3}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( \frac{9i^2}{n^2} + \frac{12i}{n} - 2 \right) \left( \frac{3}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( \frac{27i^2}{n^3} + \frac{36i}{n^2} - \frac{6}{n} \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{27i^2}{n^3} + \sum\limits_{i = 1}^n \frac{36 i }{n^2} - \sum\limits_{i = 1}^n \frac{6}{n}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{27}{n^3} \sum\limits_{i = 1}^n + \frac{36}{n^2} \sum\limits_{i = 1}^n i - \frac{1}{n} \sum\limits_{i = 1}^n 6\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{27}{n^3} \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{36}{n^2} \left( \frac{n(n+1)}{2} \right) - \frac{1}{n} (6n)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{9(n+1)(2n+1)}{2n^2} + \frac{18(n+1)}{n} - 6 \\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{18n^2 + 27 n + 9}{2n^2} + \frac{18n + 18}{n} -6 \\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{18n^2 + 27n + 9 + 36n^2 + 36n - 12n^2}{2n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \frac{42n^2 + 63n + 9}{2n^2}
\end{aligned}
\end{equation}
$

Evaluating the limit


$
\begin{equation}
\begin{aligned}
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n f(x_i) \Delta x\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \lim_{n \to \infty} \left( \frac{42n^2 + 63n + 9}{2n^2} \right)\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \lim_{n \to \infty} \left( \frac{\frac{42\cancel{n^2}}{\cancel{n^2}} + \frac{63n}{n^2} + \frac{9}{n^2}}{\frac{2\cancel{n^2}}{\cancel{n^2}}} \right)\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \lim_{n \to \infty} \left( \frac{42 + \frac{63}{n} + \frac{9}{n^2} }{2} \right)\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \frac{42 + \lim\limits_{n \to \infty}\frac{63}{n} + \lim\limits_{n \to \infty} \frac{9}{n^2} }{2}\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \frac{42+0+0}{2}\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= \frac{42}{2}\\
\\
\int^4_1 \left( x^2 + 2x - 5 \right) dx &= 21
\end{aligned}
\end{equation}
$