find the average value (g_{ave}) of the function (g) on the given interval.(g(t)=\frac{t}{sqrt{7 +…

Explanation:

Step1: Recall average - value formula

The average value of a function $y = g(t)$ on the interval $[a,b]$ is given by $g_{ave}=\frac{1}{b - a}\int_{a}^{b}g(t)dt$. Here, $a = 3$, $b = 7$, and $g(t)=\frac{t}{\sqrt{7 + t^{2}}}$. So, $g_{ave}=\frac{1}{7 - 3}\int_{3}^{7}\frac{t}{\sqrt{7 + t^{2}}}dt=\frac{1}{4}\int_{3}^{7}\frac{t}{\sqrt{7 + t^{2}}}dt$.

Step2: Use substitution

Let $u = 7 + t^{2}$, then $du=2tdt$, and $tdt=\frac{1}{2}du$. When $t = 3$, $u=7 + 3^{2}=7 + 9 = 16$. When $t = 7$, $u=7+7^{2}=7 + 49 = 56$. So, $\frac{1}{4}\int_{3}^{7}\frac{t}{\sqrt{7 + t^{2}}}dt=\frac{1}{4}\int_{16}^{56}\frac{1}{2\sqrt{u}}du=\frac{1}{8}\int_{16}^{56}u^{-\frac{1}{2}}du$.

Step3: Integrate

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-\frac{1}{2}$, we have $\int u^{-\frac{1}{2}}du = 2u^{\frac{1}{2}}+C$. Then $\frac{1}{8}\int_{16}^{56}u^{-\frac{1}{2}}du=\frac{1}{8}\left[2u^{\frac{1}{2}}\right]_{16}^{56}$.

Step4: Evaluate the definite integral

$\frac{1}{8}\times2\left[\sqrt{u}\right]_{16}^{56}=\frac{1}{4}(\sqrt{56}-\sqrt{16})=\frac{1}{4}(2\sqrt{14}-4)=\frac{\sqrt{14}-2}{2}$.

Answer:

$\frac{\sqrt{14}-2}{2}$

Now for the second function $f(t)=e^{\sin(t)}\cos(t)$ on the interval $\left[0,\frac{\pi}{2}\right]$:

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(t)$ on the interval $[a,b]$ is $f_{ave}=\frac{1}{b - a}\int_{a}^{b}f(t)dt$. Here, $a = 0$, $b=\frac{\pi}{2}$, and $f(t)=e^{\sin(t)}\cos(t)$. So, $f_{ave}=\frac{1}{\frac{\pi}{2}-0}\int_{0}^{\frac{\pi}{2}}e^{\sin(t)}\cos(t)dt=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}e^{\sin(t)}\cos(t)dt$.

Step2: Use substitution

Let $u=\sin(t)$, then $du=\cos(t)dt$. When $t = 0$, $u=\sin(0)=0$. When $t=\frac{\pi}{2}$, $u=\sin\left(\frac{\pi}{2}\right)=1$. So, $\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}e^{\sin(t)}\cos(t)dt=\frac{2}{\pi}\int_{0}^{1}e^{u}du$.

Step3: Integrate

The integral of $e^{u}$ with respect to $u$ is $e^{u}+C$. Then $\frac{2}{\pi}\int_{0}^{1}e^{u}du=\frac{2}{\pi}\left[e^{u}\right]_{0}^{1}$.

Step4: Evaluate the definite integral

$\frac{2}{\pi}(e^{1}-e^{0})=\frac{2(e - 1)}{\pi}$.

Answer:

$\frac{2(e - 1)}{\pi}$