Answer:
The Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints is about 24.328125.
Step-by-step explanation:
We want to find the Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints.
The Midpoint Sum uses the midpoints of a sub-interval:
[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We know that a = 4, b = 5, n = 4.
Therefore, [tex]\Delta{x}=\frac{5-4}{4}=\frac{1}{4}[/tex]
Divide the interval [4, 5] into n = 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{4}[/tex]
[tex]\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right][/tex]
Now, we just evaluate the function at the midpoints:
[tex]f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625[/tex]
[tex]f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625[/tex]
[tex]f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625[/tex]
[tex]f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625[/tex]
Finally, use the Midpoint Sum formula
[tex]\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125[/tex]
This is the sketch of the function and the approximating rectangles.
