Question

Approximate the area under the curve over the specified interval by using the indicated number of subintervals (or rectangles) and evaluating the function at the right-hand endpoints of the subintervals. (See Example 1.)f(x) = 9 − x2 from x = 1 to x = 3; 4 subintervals

Answer 1

Answer:

The area under the function $\int\limits^3_1 {9-x^2} \, dx \approx 7.25.$.

Step-by-step explanation:

We want to find the Riemann Sum for $\int\limits^3_1 {9-x^2} \, dx$ 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:

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_1)+f(x_2)+f(x_3)+...+f(x_{n-1})+f(x_{n})\right)$

where $\Delta{x}=\frac{b-a}{n}$

From the information given we know that a = 1, b = 3, n = 4.

Therefore, $\Delta{x}=\frac{3-1}{4}=\frac{1}{2}$

We need to divide the interval [1, 3] into 4 sub-intervals of length $\Delta{x}=\frac{1}{2}$:

$\left[1, \frac{3}{2}\right], \left[\frac{3}{2}, 2\right], \left[2, \frac{5}{2}\right], \left[\frac{5}{2}, 3\right]$

Now, we just evaluate the function at the right endpoints:

$f\left(x_{1}\right)=f\left(\frac{3}{2}\right)=\frac{27}{4}=6.75$

$f\left(x_{2}\right)=f\left(2\right)=5=5$

$f\left(x_{3}\right)=f\left(\frac{5}{2}\right)=\frac{11}{4}=2.75$

$f\left(x_{4}\right)=f(b)=f\left(3\right)=0=0$

Next, we use the Right Riemann Sum formula

$\int\limits^3_1 {9-x^2} \, dx \approx \frac{1}{2}(6.75+5+2.75+0)=7.25$