Answer with Step-by-step explanation:
We are given that
[tex]f(x)=4x[/tex]
Interval=[2,5]
[tex]h=\frac{b-a}{n}=\frac{5-2}{n}=\frac{3}{n}[/tex]
[tex]x_i=i\frac{3}{n}[/tex]
Where i=1,2,3,... n
[tex]f(x_i)=4i\times \frac{3}{n}=\frac{12i}{n}[/tex]
Riemann sum=[tex]\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(x_i)\cdot h=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}(\frac{12i}{n}\times \frac{3}{n}[/tex]
Riemann sum=[tex]\lim_{n\rightarrow \infty}\frac{36}{n^2}\sum_{i=1}^{n}i[/tex]
Riemann sum=[tex]\lim_{n\rightarrow \infty}\frac{36}{n^2}\times \frac{n(n+1)}{2}[/tex]
By using
[tex]\sum n=\frac{n(n+1)}{2}[/tex]
Riemann sum=[tex]\lim_{n\rightarrow \infty}\frac{18n(n+1)}{n^2}=\lim_{n\rightarrow \infty}18(1+\frac{1}{n})[/tex]
Apply the limit
Area under the curve=[tex]18[/tex] square units
