find the average value of the function below on the interval that is given. f(x) = 2√(2x + 1); 3, 5

Explanation:

Step1: Recall average - value formula

The average value of a function $y = f(x)$ on the interval $[a,b]$ is given by $f_{avg}=\frac{1}{b - a}\int_{a}^{b}f(x)dx$. Here, $a = 3$, $b = 5$, and $f(x)=2\sqrt{2x + 1}$. So, $f_{avg}=\frac{1}{5 - 3}\int_{3}^{5}2\sqrt{2x + 1}dx=\int_{3}^{5}\sqrt{2x + 1}dx$.

Step2: Use substitution

Let $u = 2x+1$, then $du=2dx$. When $x = 3$, $u=2\times3 + 1=7$; when $x = 5$, $u=2\times5+1 = 11$. And $\int\sqrt{2x + 1}dx=\frac{1}{2}\int u^{\frac{1}{2}}du$.

Step3: Integrate $u^{\frac{1}{2}}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{2}\int u^{\frac{1}{2}}du=\frac{1}{2}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}u^{\frac{3}{2}}+C$.

Step4: Evaluate the definite integral

$\int_{3}^{5}\sqrt{2x + 1}dx=\left[\frac{1}{3}(2x + 1)^{\frac{3}{2}}\right]_{3}^{5}=\frac{1}{3}(11)^{\frac{3}{2}}-\frac{1}{3}(7)^{\frac{3}{2}}$. [ \begin{align*} \frac{1}{3}(11)^{\frac{3}{2}}-\frac{1}{3}(7)^{\frac{3}{2}}&=\frac{1}{3}(11\sqrt{11}-7\sqrt{7})\ &=\frac{1}{3}(11\times3.3166 - 7\times2.6458)\ &=\frac{1}{3}(36.4826-18.5206)\ &=\frac{1}{3}\times17.962\ &\approx5.99 \end{align*} ]

Answer:

$5.99$