Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 26

a.) Determine the approximation to the integral $\displaystyle \int^4_0 \left( x^2 - 3x \right) dx$ using Riemann Sum with right end points and $n = 8$
Let $f(x) = x^2 - 3x$
With $n= 8$ the interval width is $\displaystyle \Delta x = \frac{b-a}{n} = \frac{4-0}{8} = \frac{4}{8} = \frac{1}{2}$

And the right end points are $x_1 = 0.5$, $x_2 = 1.0$, $x_3 = 1.5$, $x_4 = 2.0$, $x_5 = 2.5$, $x_6 = 3.0$, $x_7 = 3.5$ and $x_8 = 4.0$

So the Riemann Sum is

$
\begin{equation}
\begin{aligned}
R_8 &= \sum\limits_{i = 1}^8 f(x_i) \Delta x\\
\\
R_8 &= f(0.5) \Delta x + f(1.0) \Delta x + f(1.5) \Delta x + f(2.0) \Delta x + f(2.5) \Delta x + f(3.0) \Delta x + f(3.5) \Delta x + f(4.0) \Delta x\\
\\
R_8 &= \left( \frac{-5}{4} \right) \left( \frac{1}{2} \right) + (-2) \left( \frac{1}{2} \right) + \left( \frac{-9}{4} \right) \left( \frac{1}{2} \right) + (-2) \left( \frac{1}{2} \right) + \left( \frac{-5}{4} \right) \left( \frac{1}{2} \right) + (0) \left( \frac{1}{2} \right) + \left( \frac{7}{4} \right) \left( \frac{1}{2} \right) + (4) \left( \frac{1}{2} \right)\\
\\
R_8 &= \frac{-5}{8} - 1 - \frac{9}{8} - 1 - \frac{5}{8} + \frac{7}{8} + 2 \\
\\
R_8 &= \frac{-5-8-9-8-5+7+16}{8}\\
\\
R_8 &= \frac{-12}{8}\\
\\
R_8 &= \frac{-3}{2}\\
\\
R_8 &= -1.5
\end{aligned}
\end{equation}
$


b.) Sketch a diagram to illustrate the approximation in part(a)



Evaluate $\displaystyle \int^4_0 \left( x^2 - 3x) \right) dx$
Using the definition of the integral
$\displaystyle \int^b_a f(x) dx = \lim\limits_{n \to \infty} \sum\limits_{i = 1}^n f(x_i) \Delta x$

$
\begin{equation}
\begin{aligned}
\Delta x &= \frac{b-a}{n}\\
\\
\Delta x &= \frac{4-0}{n}\\
\\
\Delta x &= \frac{4}{n}\\
\\
x_i &= a + i \Delta x\\
\\
x_i &= 0 + \frac{4}{n} i\\
\\
x_i &= \frac{4i}{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{4i}{n} \right) \left( \frac{4}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ \left( \frac{4i}{n} \right)^2 - 3 \left( \frac{4i}{n} \right) \right] \left( \frac{4}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( \frac{16i^2}{n^2} - \frac{12i}{n} \right) \left( \frac{4}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left( \frac{64i^2}{n^3} - \frac{48i}{n^2} \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{64i^2}{n^3} - \sum\limits_{i = 1}^n \frac{48i}{n^2} \\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{64}{n^3} \sum\limits_{i = 1}^n i^2 - \frac{48}{n^2} \sum\limits_{i = 1}^n i \\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{64}{n^3} \left[ \frac{n(n+1)(2n+1)}{6} \right] - \frac{48}{n^2} \left[ \frac{n(n+1)}{2} \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{32(n+1)(2n+1)}{3n^2} - \frac{[24(n+1)]}{n}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{32(2n^2+n+2n+1)}{3n^2} - \frac{(24n+24)}{n}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{64n^2+96n+32}{3n^2} - \frac{(24n+24)}{n}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{64n^2 + 96n + 32 - 72n^2 - 72n}{3n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{-8n^2 + 24n + 32}{3n^2}
\end{aligned}
\end{equation}
$


Evaluating the limit

$
\begin{equation}
\begin{aligned}
\int^4_0 \left( x^2 - 3x \right) dx &= \lim_{n \to \infty} \sum\limits_{i = 1}^n f(x_i) \Delta x\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \lim_{n \to \infty} \left( \frac{-8n^2+24n+32}{3n^2} \right)\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \lim_{n \to \infty} \left( \frac{\frac{-8\cancel{n^2}}{\cancel{n^2}} + \frac{24n}{n}+\frac{32}{n^2} }{\frac{3\cancel{n^2}}{\cancel{n^2}}} \right)\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \lim_{n \to \infty} \left( \frac{-8 + \frac{24}{n}+\frac{32}{n} }{3} \right)\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \frac{-8 + \lim\limits_{n \to \infty}\frac{24}{n}+ \lim\limits_{n \to \infty} \frac{32}{n^2} }{3}\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \frac{-8+0+0}{3}\\
\\
\int^4_0 \left( x^2 - 3x \right) dx &= \frac{-8}{3}
\end{aligned}
\end{equation}
$