Given the definite integral to find the total area between the graph of f(x)
[tex]f(x)=3x+3\text{ on }\lbrack0,4\rbrack[/tex]We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 , b = 4
[tex]Fori=0,1,2,\ldots,n,letP=\mleft\{\xi\mright\}bearegularpartitionof\mleft[0,4\mright].Then[/tex]Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate the function value at the right endpoint of the interval [xi−1,xi]. The right endpoint of the interval is xi, and since P is a regular partition,
[tex]\begin{gathered} \Delta x=\frac{b-a}{n} \\ \Delta x=\frac{4-0}{n}=\frac{4}{n} \\ x_i=x_0+i\Delta x=0+i\frac{4}{n} \end{gathered}[/tex]Thus, the function value at the right endpoint of the interval is
[tex]\begin{gathered} f(x_i)=3(\frac{i4}{n})+3 \\ f(x_i)=\frac{12i}{n}+3 \end{gathered}[/tex]Then the Riemann sum takes the form
[tex]\sum ^n_{i=1}f(x_i)\Delta x=\sum ^n_{i=1}(\frac{12i}{n}+3_{})\Delta x[/tex]Therefore,
[tex]\sum ^n_{i=1}f(x_i)\Delta x=\sum ^n_{i=1}\frac{12i}{n}_{}\Delta x+\sum ^n_{i=1}3_{}\Delta x[/tex]Hence,
[tex]\sum ^n_{i=1}f(x_i)\Delta x=\frac{12\Delta x}{n}\sum ^n_{i=1}i_{}+3_{}\Delta x\sum ^n_{i=1}1[/tex]Thus,
[tex]\begin{gathered} \sum ^n_{i=1}f(x_i)\Delta x=\frac{12\Delta x}{n}\times\frac{n(n+1)}{2})+3_{}\Delta x\times n \\ \\ \sum ^n_{i=1}f(x_i)\Delta x=\frac{24}{n}\times\frac{(n+1)}{1})+12 \\ \sum ^n_{i=1}f(x_i)\Delta x=\frac{24(n+1)}{n})+12=24(1+\frac{1}{n})+12 \end{gathered}[/tex]Hence,
[tex]\lim _{n\to\infty}\sum ^n_{i=1}f(x_i)\Delta x=\lim _{n\to\infty}24(1+\frac{1}{n})+12=24+12=36[/tex][tex]\begin{gathered} \text{ Since }\lim _{n\to\infty}\sum ^n_{i=1}f(x_i)\Delta x=\int ^4_03x+3dx\text{ then} \\ \int ^4_03x+3dx=36 \end{gathered}[/tex]Therefore the definite integral for the function is 36
