Answer :
Answer:
The area under the function [tex]\int\limits^3_1 {9-x^2} \, dx \approx 7.25.[/tex].
Step-by-step explanation:
We want to find the Riemann Sum for [tex]\int\limits^3_1 {9-x^2} \, dx[/tex] with 4 sub-intervals, using right endpoints.
A Riemann Sum is a method for approximating the total area underneath a curve on a graph, otherwise known as an integral.
The Right Riemann Sum is given by:
[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_1)+f(x_2)+f(x_3)+...+f(x_{n-1})+f(x_{n})\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
From the information given we know that a = 1, b = 3, n = 4.
Therefore, [tex]\Delta{x}=\frac{3-1}{4}=\frac{1}{2}[/tex]
We need to divide the interval [1, 3] into 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{2}[/tex]:
[tex]\left[1, \frac{3}{2}\right], \left[\frac{3}{2}, 2\right], \left[2, \frac{5}{2}\right], \left[\frac{5}{2}, 3\right][/tex]
Now, we just evaluate the function at the right endpoints:
[tex]f\left(x_{1}\right)=f\left(\frac{3}{2}\right)=\frac{27}{4}=6.75[/tex]
[tex]f\left(x_{2}\right)=f\left(2\right)=5=5[/tex]
[tex]f\left(x_{3}\right)=f\left(\frac{5}{2}\right)=\frac{11}{4}=2.75[/tex]
[tex]f\left(x_{4}\right)=f(b)=f\left(3\right)=0=0[/tex]
Next, we use the Right Riemann Sum formula
[tex]\int\limits^3_1 {9-x^2} \, dx \approx \frac{1}{2}(6.75+5+2.75+0)=7.25[/tex]