Question:

Consider the function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$.

(a) Find the average value of the function over this interval.

(b) Determine at least one value of $c$ in $(1, 4)$ that satisfies the Mean Value Theorem for Integrals for this function on the given interval.

Answer:

(a) Average Value

To find the average value of a function $f(x)$ over an interval $[a, b]$, we need to evaluate the definite integral of $f(x)$ over that interval and then divide the result by the length of the interval.

The average value of $f(x)$ over the interval $[a, b]$ is given by the formula:

For the given function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$, we can find the average value as follows:

\begin{align*} \text{Average Value} &= \frac{1}{4-1} \int_1^4 (3x^2 + 2x - 4) ,dx \ &= \frac{1}{3} \int_1^4 (3x^2 + 2x - 4) ,dx \ &= \frac{1}{3}\left[\frac{3x^3}{3} + \frac{2x^2}{2} - 4x\right]_1^4 \ &= \frac{1}{3}\left[x^3 + x^2 - 4x\right]_1^4 \ &= \frac{1}{3}\left[(4^3 + 4^2 - 4(4)) - (1^3 + 1^2 - 4(1))\right] \ &= \frac{1}{3}\left[64 + 16 - 16 - 1 - 1 + 4\right] \ &= \frac{1}{3}\left[66\right] \ &= \boxed{22} \end{align*}

Therefore, the average value of the function $f(x)$ over the interval $[1, 4]$ is $22$.

(b) Mean Value Theorem for Integrals

According to the Mean Value Theorem for Integrals, if a function $f(x)$ is continuous on the closed interval $[a, b]$, then there exists at least one value $c$ in $(a, b)$ such that the instantaneous rate of change of $f(x)$ at $c$ is equal to the average rate of change of $f(x)$ over $[a, b]$.

For the given function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$, we need to find a value $c$ in $(1, 4)$ that satisfies the Mean Value Theorem for Integrals.

First, let's find the average rate of change of $f(x)$ over $[1, 4]$. To do this, we calculate the derivative of $f(x)$ and then evaluate it at some value in $(1, 4)$.

\begin{align*} f(x) &= 3x^2 + 2x - 4 \ f'(x) &= 6x + 2 \end{align*}

Now, we evaluate $f'(x)$ at $c$, leading to:

\begin{align*} f'(c) &= 6c + 2 \end{align*}

Next, we need to find the average rate of change of $f(x)$ over $[1, 4]$. To do this, we calculate $\frac{f(b) - f(a)}{b - a}$, where $a = 1$ and $b = 4$.

\begin{align*} \frac{f(b) - f(a)}{b - a} &= \frac{f(4) - f(1)}{4 - 1} \ &= \frac{(3(4)^2 + 2(4) - 4) - (3(1)^2 + 2(1) - 4)}{4 - 1} \ &= \frac{(48 + 8 - 4) - (3 + 2 - 4)}{3} \ &= \frac{52 - 5}{3} \ &= \frac{47}{3} \end{align*}

To satisfy the Mean Value Theorem for Integrals, we need to find $c$ in $(1, 4)$ such that $f'(c) = \frac{47}{3}$.

Setting up the equation:

\begin{align*} 6c + 2 &= \frac{47}{3} \ 6c &= \frac{47}{3} - 2 \ 6c &= \frac{47 - 6 \times 2}{3} \ 6c &= \frac{47 - 12}{3} \ 6c &= \frac{35}{3} \ c &= \frac{35}{18} \end{align*}

Therefore, there exists at least one value $c$ in $(1, 4)$ (namely $c = \frac{35}{18}$) that satisfies the Mean Value Theorem for Integrals for the function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$.

In summary:

(a) The average value of the function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$ is $22$.

(b) The value of $c$ that satisfies the Mean Value Theorem for Integrals for the function $f(x) = 3x^2 + 2x - 4$ over the interval $[1, 4]$ is $c = \frac{35}{18}$.