009B Sample Midterm 1, Problem 2

Find the average value of the function on the given interval.

$f(x)=2x^{3}(1+x^{2})^{4},~~~[0,2]$

Foundations:

The average value of a function $f(x)$ on an interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} is given by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)dx} .

Solution:

Step 1:
Using the formula given in the Foundations sections, we have:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_{\text{avg}}=\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4dx=\int_0^2 x^3(1+x^2)^4dx}

Step 2:
Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1+x^2)^2=1+2x^2+x^4} , we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1+x^2)^4=(1+x^2)^2(1+x^2)^2=(1+2x^2+x^4)(1+2x^2+x^4)=1+4x^2+6x^4+4x^6+x^8} .

Final Answer:

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