The Riemann Sum Definite Integral Formula,
[tex]\int ^b_af(x)dx=\lim _{n\to\infty}\sum ^n_{i\mathop=1}f(x^{}_{i^{\ast}})\Delta x[/tex]Where
[tex]\begin{gathered} \Delta x=\frac{b-a}{n} \\ x_{i^{\ast}}=a+(\Delta x)i \\ \end{gathered}[/tex]and,
[tex]\begin{gathered} \sum ^n_{i\mathop=1}k=kn \\ \sum ^n_{i\mathop{=}1}i=\frac{n(n+1)}{2} \\ \sum ^n_{i\mathop{=}1}i^2=\frac{n(n+1)(2n+1)}{6} \end{gathered}[/tex]The integral to evaluate is
[tex]\int ^b_ax^2dx[/tex]Lets find x_i *:
[tex]\begin{gathered} x_{i^{\ast}}=a+(\Delta x)i \\ x_{i^{\ast}}=a+(\frac{b-a}{n})i \\ x_{i^{\ast}}=a+\frac{(b-a)i}{n} \\ x_{i^{\ast}}=a+\frac{bi-ai}{n} \\ x_{i^{\ast}}=\frac{an+bi-ai}{n} \end{gathered}[/tex]Now, let's find f(x_i*):
[tex]f(x_{i^{\ast}})=(\frac{an+bi-ai}{n})^2=(\frac{an+bi-ai}{n})(\frac{an+bi-ai}{n})=\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2}[/tex]Now, putting it into the formula,
[tex]\begin{gathered} \int ^b_af(x)dx=\lim _{n\to\infty}\sum ^n_{i\mathop{=}1}f(x^{}_{i^{\ast}})\Delta x \\ \lim _{n\to\infty}\sum ^n_{i\mathop{=}1}\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2}\times\frac{b-a}{n} \end{gathered}[/tex]Simplifying:
[tex]\begin{gathered} \lim _{n\to\infty}\sum ^n_{i\mathop{=}1}(\frac{a^2n^2+2\text{anbi}-2a^2ni-2\text{abi}^2+a^2i^2+b^2i^2}{n^2})\times\frac{b-a}{n} \\ \lim _{n\to\infty}\sum ^n_{i\mathop{=}1}(a^2+\frac{2\text{abi}}{n}-\frac{2a^2i}{n}-\frac{2\text{abi}^2}{n^2}+\frac{a^2i^2}{n^2}+\frac{b^2i^2}{n^2})\times\frac{b-a}{n} \\ \lim _{n\to\infty}(\frac{b-a}{n})\lbrack\sum ^n_{i\mathop{=}1}a^2+\sum ^n_{i\mathop{=}1}\frac{2\text{abi}}{n}-\sum ^n_{i\mathop{=}1}\frac{2a^2i}{n}-\sum ^n_{i\mathop{=}1}\frac{2\text{abi}^2}{n^2}+\sum ^n_{i\mathop{=}1}\frac{a^2i^2}{n^2}+\sum ^n_{i\mathop{=}1}\frac{b^2i^2}{n^2}\rbrack \end{gathered}[/tex]Further simplifying:
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