Question

In Exercises \(11 - 14 ,\) use NINT to find the average value of the function on the interval. At what point (s) in the interval does the function assume its average value? $$y = x ^ { 2 } - 1 , [ 0 , \sqrt { 3 } ]$$

Short answer

The average value of the function is calculated to find the points where the function assume its average value.

Step-by-step solution

Find the average value of the function

To find the average value of the function \(f(x) = x^2 - 1\) over the interval \([0, \sqrt{3}]\), we can use the formula for the average value of a function on a closed interval \([a, b]\): \[ \bar{f} = \frac{1}{b - a} \int_{a}^{b} f(x) dx \]Where \(a\) and \(b\) are the limits of the interval. Here, \(a = 0\), \(b = \sqrt{3}\), and \(f(x) = x^2 - 1\). Therefore, the average value \(\bar{f}\) is calculated as follows: \[ \bar{f} = \frac{1}{\sqrt{3} - 0} \int_{0}^{\sqrt{3}} (x^2 - 1) dx \]Now we need to calculate the definite integral and multiply it by \(\frac{1}{\sqrt{3}}\).

Calculate the definite integral

The integral can be calculated using the power rule for integration, which states that \[ \int x^n dx = \frac{x^{n+1}}{n+1} \]So, we get \[ \int_{0}^{\sqrt{3}} (x^2 - 1) dx = \left[\frac{x^{2+1}}{2+1} -x \right]_0^{\sqrt{3}} \]Calculating this gives us the definite integral, which we then multiply by \(\frac{1}{\sqrt{3}}\).

Find the points where the function equals its average value

To find the points where the function \(f(x) = x^2 - 1\) assumes its average value (the value we calculated in step 2), we need to solve the equation \(f(x) = \bar{f}\). This will yield the x-coordinates of the points.