Question

find the average value $f_{ave}$ of the function $f$ on the given interval.
$f(x)=3x^{2}+4x,-1,5$
$f_{ave}=
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find the average value $g_{ave}$ of the function $g$ on the given interval.
$g(x)=7cos(x),-\frac{pi}{2},\frac{pi}{2}$
$g_{ave}=$
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Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{ave}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$.

Step2: Calculate $f_{ave}$ for $f(x)=3x^{2}+4x$ on $[-1,5]$

First, find $\int_{-1}^{5}(3x^{2}+4x)dx$.
Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have:
$\int(3x^{2}+4x)dx=3\times\frac{x^{3}}{3}+4\times\frac{x^{2}}{2}+C=x^{3}+2x^{2}+C$.
Then, evaluate the definite integral:
$\int_{-1}^{5}(3x^{2}+4x)dx=(x^{3}+2x^{2})\big|_{-1}^{5}=(5^{3}+2\times5^{2})-((-1)^{3}+2\times(-1)^{2})$.
$=(125 + 50)-(-1 + 2)=175-1 = 174$.
Since $a=-1$ and $b = 5$, $b - a=6$.
So, $f_{ave}=\frac{1}{6}\times174 = 29$.

Step3: Calculate $g_{ave}$ for $g(x)=7\cos(x)$ on $[-\frac{\pi}{2},\frac{\pi}{2}]$

First, find $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}7\cos(x)dx$.
Since $\int\cos(x)dx=\sin(x)+C$, then $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}7\cos(x)dx=7\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)dx$.
Using the property $\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx$ for an even function $f(x)$ (and $\cos(x)$ is an even function), we have $7\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(x)dx=7\times2\int_{0}^{\frac{\pi}{2}}\cos(x)dx$.
$=14\times\sin(x)\big|_{0}^{\frac{\pi}{2}}=14\times(1 - 0)=14$.
Since $a =-\frac{\pi}{2}$ and $b=\frac{\pi}{2}$, $b - a=\pi$.
So, $g_{ave}=\frac{14}{\pi}$.

Answer:

$f_{ave}=29$
$g_{ave}=\frac{14}{\pi}$