Answer:
The average value of g is:
[tex]\displaystyle g_{ave}=\frac{1}{6}e^6-\frac{49}{6}\approx 59.071[/tex]
Step-by-step explanation:
The average value of a function is given by the formula:
[tex]\displaystyle f_{ave}=\frac{1}{b-a}\int_a^b f(x)\, dx[/tex]
We want to find the average value of the function:
[tex]g(x)=e^{3x-3}-4x[/tex]
On the interval [1, 3].
So, the average value will be given by:
[tex]g_{ave}=\displaystyle \frac{1}{3-1}\int_1^3 e^{3x-3}-4x\, dx[/tex]
Simplify. We will also split the integral:
[tex]\displaystyle g_{ave}=\frac{1}{2}\left(\int_1^3e^{3x-3}\, dx-\int _1^3 4x\, dx\right)[/tex]
We can use u-substitution for the first integral. Letting u = 3x - 3, we acquire:
[tex]\displaystyle u=3x-3\Rightarrow du = 3\, dx\Rightarrow \frac{1}{3} du=dx[/tex]
We will also change the limits of integration for our first integral. So:
[tex]u(1)=3(1)-3=0\text{ and } u(3)=3(3)-3=6[/tex]
Thus:
[tex]\displaystyle g_{ave}=\frac{1}{2}\left(\frac{1}{3}\int_0^6 e^{u}\, du-\int _1^3 4x\, dx\right)[/tex]
Integrate:
[tex]g_{ave}=\displaystyle \frac{1}{2}\left(\frac{1}{3}e^u\Big|_0^6-2x^2\Big|_1^3\right)[/tex]
Evaluate. So, the average value of g on the interval [1, 3] is:
[tex]\displaystyle g_{ave}=\frac{1}{2}\left(\frac{1}{3}\left[e^6-e^0\right]-\left[2(3)^2-2(1)^2\right]\right)[/tex]
Evaluate:
[tex]\displaystyle\begin{aligned} g_{ave}&=\frac{1}{2}\left(\frac{1}{3}(e^6-1)-16\right)\\&=\frac{1}{2}\left(\frac{1}{3}e^6-\frac{1}{3}-16\right)\\&=\frac{1}{6}e^6-\frac{49}{6}\approx59.071\end{aligned}[/tex]
