When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation is very helpful because it provides a simple way to determine the average value by substituting in values of the bounds and the function itself:

\(\displaystyle f_{avg}=\frac{1}{5-0}\int_{0}^{5}4e^{2x} dx=(\frac{1}{5})[4e^{2x} * \frac{1}{2}] |_0^5 =\frac{2}{5}e^{2(5)}-\frac{2}{5}e^{2(0)}=\frac{2}{5}(e^{10}-1)\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of t, so the definite integral expression should likewise be in terms of \(\displaystyle t\).

\(\displaystyle f_{avg}=\frac{1}{\frac{\pi}{4}-0}\int_{0}^{\frac{\pi}{4}}4+2cos(2t) dt=(\frac{4}{\pi})[4t+sin(2t)] |_0^{\frac{\pi}{4}}\)\(\displaystyle =(\frac{4}{\pi})[\pi +1-0-0]=4+\frac{4}{\pi}\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function and interval to solve for the average value. While the function in this problem contains a trigonometric function, the same approach can be applied. Remember that the function is in terms of \(\displaystyle t\), so the definite integral expression should likewise be in terms of \(\displaystyle t\).

\(\displaystyle f_{avg}=\frac{1}{\pi-0}\int_{0}^{\pi}6-(3t)sin(t^2) dt =\frac{1}{\pi}\int_{0}^{\pi}6-3tsin(t^2) dt\)

Because the objective of this problem is to find the average value of the function, the formula f\(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\) will be useful. Since the interval and function are provided, this problem consists of recognizing the base components and making the appropriate substitutions:

\(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx=\frac{1}{2-1}\int_{1}^{2}2x^5-6x^3+7 dx =[\frac{1}{3}x^6-\frac{3}{2}x^4+7x] |_1^2\)\(\displaystyle =\frac{64}{3}-24+14-\frac{1}{3}+\frac{3}{2}-7=5.5\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function and interval to solve for the average value. Because the function indicated in the problem is in terms of \(\displaystyle x\), the definite integral expression should also be in terms of \(\displaystyle x\).

\(\displaystyle f_{avg}=\frac{1}{4-1}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx =\frac{1}{3}\int_{1}^{4}\frac{1}{x}+\sqrt{2x} dx\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function and interval to solve for the average value. Because the function indicated in the problem is in terms of \(\displaystyle x\), the definite integral expression should also be in terms of \(\displaystyle x\).

\(\displaystyle f_{avg}=\frac{1}{1-(-1)}\int_{-1}^{1}20x+x^3 dx\)\(\displaystyle =\frac{1}{2}[10x^2+(\frac{1}{4})(x)^4] |_-1^1=\frac{1}{2}(10+\frac{1}{4}-10-\frac{1}{4})=0\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function, average value, and interval to solve for \(\displaystyle c\).

\(\displaystyle 27=\frac{1}{c-0}\int_{c}^{0}4x +11 dx\)

Next, the definite integral can be taken to continue solving for \(\displaystyle c\).

\(\displaystyle 27=\frac{1}{c}[2x^2+11x] |_0^c\)

\(\displaystyle 27=(\frac{1}{c})(2c^2+11c)\)

\(\displaystyle 27=2c+11\)

\(\displaystyle c=8\)

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation allows the substitution of the function, average value, and interval to solve for \(\displaystyle c\).

\(\displaystyle 12=\frac{1}{c-0}\int_{0}^{c}3x^2+4x-3 dx\)

Next, the definite integral can be taken to continue solving for \(\displaystyle c\).

\(\displaystyle 12=\frac{1}{c}[x^3+2x^2-3x] |_0^c\)

\(\displaystyle 12=(\frac{1}{c})(c^3+2c^2-3c)\)

\(\displaystyle 12=c^2+2c-3\)

\(\displaystyle 0=c^2+2c-15\)

\(\displaystyle 0=(c+5)(c-3)\)

\(\displaystyle c={-5,3}\)

Because the problem states that \(\displaystyle c>0\), the answer \(\displaystyle c=-5\) can be eliminated. Therefore, the correct answer is \(\displaystyle c=3\).

The following equation is used for finding the Average Value of a Function:  \(\displaystyle f_{avg}=\frac{1}{b-a} \int_{a}^{b} f(x) dx\). A rearrangement of this equation could be multiplying \(\displaystyle (b-a)\) to both sides. Making this rearrangement, and substituting \(\displaystyle f_{avg}=f(c)\) with \(\displaystyle a< c< b\), results in the following: \(\displaystyle \int_{a}^{b}f(x) dx=f(c)(b-a)\). Assuming \(\displaystyle f(x)\) is continuous, this is the correct equation for the Mean Value Theorem for Integrals.

When finding the average value of a function, it is useful to keep the following formula in mind: \(\displaystyle f_{avg}=\frac{1}{b-a}\int_{a}^{b}f(x) dx\). This equation is very helpful because it provides a simple way to determine the average value by substituting in values of the bounds and the function itself:

\(\displaystyle f_{avg}=\frac{1}{3-1}\int_{1}^{3}2x^7+5x^2+4 dx=(12)[\frac{1}{4}x^8+\frac{5}{3}x^3+4x] |_1^3\)\(\displaystyle =(\frac{1}{2})(\frac{6561}{4}+45+12-1\frac{}{4}-\frac{5}{3}-4)=845.67\)