consider the given function and the given interval.\n$f(x)=(x - 5)^2$, $4,7$\n(a) find the average value…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{ave}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 4$, $b = 7$ and $f(x)=(x - 5)^2=x^{2}-10x + 25$.

Step2: Calculate the integral

First, find $\int_{4}^{7}(x^{2}-10x + 25)dx=\left[\frac{x^{3}}{3}-5x^{2}+25x\right]_{4}^{7}$. [ \begin{align*} &\left(\frac{7^{3}}{3}-5\times7^{2}+25\times7\right)-\left(\frac{4^{3}}{3}-5\times4^{2}+25\times4\right)\ =&\left(\frac{343}{3}-245 + 175\right)-\left(\frac{64}{3}-80 + 100\right)\ =&\left(\frac{343}{3}-70\right)-\left(\frac{64}{3}+20\right)\ =&\frac{343}{3}-70-\frac{64}{3}-20\ =&\frac{343 - 64}{3}-90\ =&\frac{279}{3}-90\ =&93 - 90\ =&3 \end{align*} ]

Step3: Find the average value

Since $b - a=7 - 4 = 3$, then $f_{ave}=\frac{1}{3}\times3 = 1$.

Step4: Solve for $c$

Set $f(c)=(c - 5)^2=f_{ave}=1$. Then $(c - 5)^2=1$. Taking the square - root of both sides, we get $c-5=\pm1$. If $c - 5 = 1$, then $c=6$; if $c - 5=-1$, then $c = 4$.

Answer:

(a) $f_{ave}=1$ (b) $c = 4,6$ (c) To sketch the graph of $y=(x - 5)^2$, it is a parabola with vertex at $(5,0)$. The rectangle has height $f_{ave}=1$ and base from $x = 4$ to $x = 7$. The area of the rectangle is $A=(7 - 4)\times1=3$, and the area under the curve $y=(x - 5)^2$ from $x = 4$ to $x = 7$ is also 3.