The definite integral expressed as Riemann sum is
[tex]\int ^b_af(x)dx=\lim _{n\rightarrow\infty}\sum ^n_{i=1}f(x_i)\Delta x,\text{ where }\Delta x=\frac{b-a}{n},\text{ and }x_i=a+i\Delta x[/tex]
If we want to find the graph of the function f(x) from x = 1 up to x = 5, then its definite integral is
[tex]\begin{gathered} \int ^b_af(x)dx \\ \\ \text{Substitute} \\ f(x)=x^2 \\ a=1,b=5 \\ \\ \text{THEN} \\ \int ^5_1(x^2)dx \end{gathered}[/tex]
Converting that definite integral into Riemann sum we have
[tex]\begin{gathered} \int ^5_1(x^2)dx=\lim _{n\rightarrow\infty}\sum ^n_{i=1}(1+i\frac{5-1}{n})^2(\frac{5-1}{n}) \\ =\lim _{n\rightarrow\infty}\sum ^n_{i=1}(1+i\frac{4}{n})^2\frac{4}{n} \\ =\lim _{n\rightarrow\infty}\sum ^n_{i=1}(1+\frac{4i}{n})^2\frac{4}{n} \end{gathered}[/tex][tex]\begin{gathered} \text{Therefore, the area under the graph of the function }f(x)=x^2\text{ from }x=1\text{ to }x=5 \\ \text{is the Riemann Sum} \\ \lim _{n\rightarrow\infty}\sum ^n_{i=1}(1+\frac{4i}{n})^2\frac{4}{n} \end{gathered}[/tex]
