Answer to Question #261595 in Calculus for Jess

Question #261595

Find the Mean value of the function f left parenthesis x right parenthesis space equals space x cubed space minus space 3 x squared over the interval [0, 5].

1
Expert's answer
2021-11-08T06:11:03-0500

Let us find the Mean value of the function f(x)=x33x2f(x)=x^3 - 3 x^2 over the interval [0,5].[0, 5].

It follows that the Mean value of the function is

f=15005f(x)dx=1505f(x33x2)dx=15(x44x3)05=15(6254125)=151254=254=6.25\overline{f}=\frac{1}{5-0}\int\limits_0^5f(x)dx =\frac{1}{5}\int\limits_0^5f(x^3-3x^2)dx =\frac{1}{5}(\frac{x^4}4-x^3)|_0^5\\ =\frac{1}5(\frac{625}4-125)=\frac{1}5\cdot\frac{125}4=\frac{25}{4}=6.25


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