Question

(1 point) (a) find the average value of ( f(x) = 25 - x^2 ) on the interval (0, 1). answer. (b) find a value ( c ) in the interval (0, 1) such that ( f(c) ) is equal to the average value. answer.

Explanation:

Part (a)

Step1: Recall the average value formula

The average value of a function \( f(x) \) on the interval \([a, b]\) is given by \( f_{\text{avg}}=\frac{1}{b - a}\int_{a}^{b}f(x)dx \). Here, \( a = 0 \), \( b = 1 \), and \( f(x)=25 - x^{2} \).

Step2: Compute the integral

First, find the integral of \( f(x) \) from 0 to 1: \( \int_{0}^{1}(25 - x^{2})dx=\int_{0}^{1}25dx-\int_{0}^{1}x^{2}dx \).
We know that \( \int_{0}^{1}25dx=25x\big|_{0}^{1}=25(1 - 0) = 25 \) and \( \int_{0}^{1}x^{2}dx=\frac{x^{3}}{3}\big|_{0}^{1}=\frac{1}{3}-0=\frac{1}{3} \).
So, \( \int_{0}^{1}(25 - x^{2})dx=25-\frac{1}{3}=\frac{75 - 1}{3}=\frac{74}{3} \).

Step3: Calculate the average value

Using the average value formula: \( f_{\text{avg}}=\frac{1}{1 - 0}\times\frac{74}{3}=\frac{74}{3}\approx24.67 \) (or keep it as a fraction \( \frac{74}{3} \)).

Step1: Set up the equation

We want to find \( c \) in \([0, 1]\) such that \( f(c)=f_{\text{avg}} \). We know \( f(x)=25 - x^{2} \) and \( f_{\text{avg}}=\frac{74}{3} \), so we set \( 25 - c^{2}=\frac{74}{3} \).

Step2: Solve for \( c \)

First, rearrange the equation: \( c^{2}=25-\frac{74}{3} \).
Calculate \( 25=\frac{75}{3} \), so \( c^{2}=\frac{75 - 74}{3}=\frac{1}{3} \).
Then, take the square root of both sides. Since \( c\in[0, 1] \), we take the positive root: \( c=\sqrt{\frac{1}{3}}=\frac{\sqrt{3}}{3}\approx0.577 \).

Answer:

The average value is \( \frac{74}{3} \) (or approximately \( 24.67 \))

Part (b)