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Question: How do you use the integral formula to find the average value of the function \(f\left( x \right)=18...

How do you use the integral formula to find the average value of the function f(x)=18xf\left( x \right)=18x over the interval between 00 to 44?

Explanation

Solution

In this problem we need to calculate the average value of the given function in the given range using the integral formula. We know that the average value of the function f(x)f\left( x \right) over the given limits aa and bb is given by the formula 1baabf(x)dx\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)dx}. In this formula we have the definite integral value of the function. So, we will first calculate the indefinite integral value of the given function by using the integration formulas. After that we will apply the given limits to calculate the definite integral value. Now we will apply this value in the average formula and simplify the obtained equation to get the required solution.

Complete step by step answer:
Given function f(x)=18xf\left( x \right)=18x.
Given limits are 00 and 44.
Integrating the given function to calculate the indefinite integral value, then we will get
f(x)dx=18xdx f(x)dx=18xdx \begin{aligned} & \Rightarrow \int{f\left( x \right)dx}=\int{18xdx} \\\ & \Rightarrow \int{f\left( x \right)dx}=18\int{xdx} \\\ \end{aligned}
Applying the integration formula xndx=xn+1n+1+C\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+C in the above equation, then we will have
f(x)dx=18×x22+C f(x)dx=9x2+C \begin{aligned} & \Rightarrow \int{f\left( x \right)dx}=18\times \dfrac{{{x}^{2}}}{2}+C \\\ & \Rightarrow \int{f\left( x \right)dx}=9{{x}^{2}}+C \\\ \end{aligned}
Given limits are 00 and 44.
Applying the given limits to the above calculated indefinite integral value to find the definite integral value, then we will get
04f(x)dx=9[x2]04 04f(x)dx=9(4202) 04f(x)dx=144 \begin{aligned} & \Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=9\left[ {{x}^{2}} \right]_{0}^{4} \\\ & \Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=9\left( {{4}^{2}}-{{0}^{2}} \right) \\\ & \Rightarrow \int\limits_{0}^{4}{f\left( x \right)dx}=144 \\\ \end{aligned}
Now the average value of the given function from the integral formula is given by
avg=1baabf(x)dx avg=14004f(x)dx \begin{aligned} & \Rightarrow avg=\dfrac{1}{b-a}\int\limits_{a}^{b}{f\left( x \right)dx} \\\ & \Rightarrow avg=\dfrac{1}{4-0}\int\limits_{0}^{4}{f\left( x \right)dx} \\\ \end{aligned}
Substituting the value 04f(x)dx=144\int\limits_{0}^{4}{f\left( x \right)dx}=144 in the above equation, then we will get
avg=14×144 avg=36 \begin{aligned} & \Rightarrow avg=\dfrac{1}{4}\times 144 \\\ & \Rightarrow avg=36 \\\ \end{aligned}
Hence the average value of the given function in the given limits is 3636.

Note:
In this problem we have the simple function so we can directly use the integration formula to calculate the average value without calculating the values of indefinite and definite integral values. If they have given the complex or composite functions then the above-mentioned method is very useful and gives an error less result.