Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 28

Show that $\displaystyle \int^b_a x^2 dx = \frac{b^3 - a^3}{3}$
Using the defintion of integral.
$\displaystyle \int^b_a f(x_i) dx = \lim\limits_{n \to \infty} \sum\limits_{i = 1}^2 f(x_i) \Delta x$

$
\begin{equation}
\begin{aligned}
\Delta x &= \frac{b-a}{n}\\
\\
x_i &= a + \left( \frac{b-a}{n} \right) i
\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( a + \frac{bi-ai}{n} \right) \left( \frac{b-a}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ a + \frac{(b-a)i}{n} \right]^2 \left( \frac{b-a}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ a^2 + 2ai \left( \frac{b-a}{n} \right) + \left( \frac{bi-ai}{n} \right)^2 \right] \left( \frac{b-a}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ a^2 + \frac{2abi-2a^2i}{n} + \frac{b^2i^2 - 2abi^2+a^2i^2}{n^2} \right] \left( \frac{b-a}{n} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \sum\limits_{i = 1}^n \left[ \frac{a^2(b-a)}{n} + \frac{(2abi - 2a^2i)(b-a)}{n^2} + \frac{(b^2-i^2-2abi^2 + a^2 i^2)(b-a)}{n^3} \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{a^2(b-a)}{n} + \sum\limits_{i = 1}^n \frac{\left(2ab-2a^2\right)(b-a)i}{n^2} + \sum\limits_{i = 1}^n \frac{\left(b^2 - 2ab + a^2 \right)(b-a) i^2}{n^3}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{1}{n} \sum\limits_{i = 1}^n a^2(b-a) + \frac{\left(2ab - 2a^2\right)(b-a)}{n^2} \sum\limits_{i = 1}^n i + \frac{\left(b^2 - 2ab + a^2 \right)(b-a)}{n^3} \sum\limits_{i = 1}^n i^2\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{1}{\cancel{n}} \left[a^2(b-a) \right] (\cancel{n}) + \frac{(2ab-2a^2)(b-a)}{n^2} \left[ \frac{n(n+1)}{2} \right] + \frac{\left(b^2 - 2ab + a^2\right)(b-a)}{n^3} \left[ \frac{n(n+1)(2n+1)}{6} \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= a^2(b-a) + \frac{(2ab-2a^2)(b-a)(n+1)}{2n} + \frac{\left(b^2-2ab+a^2\right)(b-a)(n+1)(2n+1)}{6n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \left[ a^2 + \frac{(2ab-2a^2)(n+1)}{2n} + \frac{\left(b^2 - 2ab + a^2\right)(n+1)(2n+1) }{6n^2} \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left[ a^2 + (n+1) \left[ \frac{2ab-2a^2}{2n} + \frac{\left(b^2-2ab+a^2\right)(2n+1}{6n^2} \right] \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left[ a^2 + (n+1) \left[ \frac{3n(2ab-2a^2)+\left(2b^2n - 4abn +2a^2n+b^2-2ab+a^2 \right)}{6n^2}\right]\right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left[ a^2 + (n+1)\left( \frac{6abn-6a^2n + 2b^2n - 4abn + 2a^2n + b^2 - 2ab + a^2}{6n^2} \right) \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left[ a^2 + (n+1) \left( \frac{2abn-4a^2n+2b^2 n+ b^2 - 2ab + a^2}{6n^2} \right) \right]\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left( \frac{6a^2 n^2 + 2abn^2 - 4a^2 n^2 + 2b^2 n^2 + b^2n - \cancel{2abn} + a^2n + \cancel{2abn} - 4a^2 n + 2b^2 n + b^2 - 2ab + a^2}{6n^2} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= (b-a) \left( \frac{2a^2 n^2 + 2ab n^2 + 2b^2 n^2 + 3b^2n - 3a^2 n + b^2 -2 ab + a^2}{6n^2} \right)\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{\cancel{2a^2 bn^2} + \cancel{2ab^2n^2}+2b^3n^2 + 3b^3n - 3a^2bn + b^3 - 2ab^2 + a^2b - 2a^3n^2 - \cancel{2a^2bn^2} - \cancel{2ab^2n^2} - 3ab^2n + 3a^3n + ab^2 + 2a^2b - a^3 }{6n^2}\\
\\
\sum\limits_{i = 1}^n f(x_i) \Delta x &= \frac{2b^3n^2 + 3b^3n - 3a^2bn+b^3-3ab^2+3a^b-2a^3n^2-3ab^2n+3a^3n-ab^2+2a^b -a^3}{6n^2}
\end{aligned}
\end{equation}
$


Evaluate the limit

$
\begin{equation}
\begin{aligned}
\int^b_a x^2 dx &= \lim_{n \to \infty} \left( \frac{2b^3n^2 + 3b^3n - 3a^2bn+b^3-3ab^2+3a^b-2a^3n^2-3ab^2n+3a^3n-ab^2+2a^b -a^3}{6n^2} \right)\\
\\
\int^b_a x^2 dx &= \lim_{n \to \infty} \left( \frac{\frac{2b^3\cancel{n^2}}{\cancel{n^2}} + \frac{3b^3n}{n^2} - \frac{3a^2bn}{n^2} + \frac{b^3}{n^2} - \frac{3ab^2}{n^2} + \frac{3a^2b}{n^2} - \frac{2a^3\cancel{n^2}}{\cancel{n^2}} - \frac{3ab^2n}{n^2} + \frac{3a^3n}{n^2} - \frac{ab^2}{n^2} + \frac{2a^2b}{n^2} - \frac{a^3}{n^2} }{\frac{6\cancel{n^2}}{\cancel{n^2}}} \right)\\
\\
\int^b_a x^2 dx &= \lim_{n \to \infty} \left( \frac{2b^3 + \frac{3b^3}{n} - \frac{3a^2b}{n} + \frac{b^3}{n^3} - \frac{3ab^2}{n^2} + \frac{3a^2b}{n^2}-2a^3-\frac{3ab^2}{n}+\frac{3a^3}{n}-\frac{ab^2}{n^2}+\frac{2a^2b}{n^2}-\frac{a^3}{n^2} }{6} \right)\\
\\
\int^b_a x^2 dx &= \frac{2b^3 + \lim\limits_{n \to \infty} \frac{3b^3}{n} - \lim\limits_{n \to \infty} \frac{3a^2b}{n} +\lim\limits_{n \to \infty} \frac{b^3}{n^2} - \lim\limits_{n \to \infty} \frac{3ab^2}{n^2} - 2a^3 - \lim\limits_{n \to \infty} \frac{3ab^2}{n} + \lim\limits_{n \to \infty} \frac{3a^3}{n} - \lim\limits_{n \to \infty} \frac{ab^2}{n^2} + \lim\limits_{n \to \infty} \frac{2a^2b}{n^2} - \lim\limits_{n \to \infty} \frac{a^3}{n^2} }{6}\\
\\
\int^b_a x^2 dx &= \frac{2b^3 + 0 - 0 + 0 - 0 - 2a^3 - 0 + 0 - 0 + 0 - 0}{6}\\
\\
\int^b_a x^2 dx &= \frac{2b^3 - 2a^3}{6}\\
\\
\int^b_a x^2 dx &= \frac{2(b^3 - a^3)}{6}\\
\\
\int^b_a x^2 dx &= \frac{b^3 - a^3}{3}
\end{aligned}
\end{equation}
$