What is the average value of the function $f (x) = 2 x^{3} {(1 + x^{2})}^{4}$ $f(x) = 2x^3(1+x^2)^4$ on the interval $[0, 2]$ $[0,2]$ ?

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What is the average value of the function $f (x) = 2 x^{3} {(1 + x^{2})}^{4}$ $f(x) = 2x^3(1+x^2)^4$ on the interval $[0, 2]$ $[0,2]$ ?

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Answer 1 · 2016-04-11T18:43:05

The average value is $\frac{4948}{5} = 989.6$ $4948/5 = 989.6$

Explanation:

The average value of $f$ $f$ on interval $[a, b]$ $[a,b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ $1/(b-a) int_a^b f(x) dx$

So we get:

$\frac{1}{2 - 0} \int_{0}^{2} 2 x^{3} {(x^{2} + 1)}^{4} d x = \frac{2}{2} \int_{0}^{2} x^{3} (x^{8} + 4 x^{6} + 10 x^{4} + 4 x^{2} + 1) d x$ $1/(2-0)int_0^2 2x^3 (x^2+1)^4 dx = 2/2 int_0^2 x^3(x^8+4x^6+10x^4+4x^2+1) dx$

$= \int_{0}^{2} (x^{11} + 4 x^{9} + 10 x^{7} + 4 x^{5} + x^{3}) d x$ $= int_0^2 (x^11+4x^9+10x^7+4x^5+x^3)dx$

$= \frac{x^{12}}{12} + \frac{4 x^{10}}{10} + \frac{6 x^{8}}{8} + \frac{4 x^{6}}{6} + \frac{x^{4}}{4}]_{0}^{2}$ $= x^12/12+(4x^10)/10 + (6x^8)/8+(4x^6)/6+x^4/4]_0^2$

$= \frac{{(2)}^{12}}{12} + \frac{2 {(2)}^{10}}{5} + \frac{3 {(2)}^{8}}{4} + \frac{2 {(2)}^{6}}{3} + \frac{{(2)}^{4}}{4}$ $= (2)^12/12+(2(2)^10)/5 + (3(2)^8)/4+(2(2)^6)/3+(2)^4/4$

$= \frac{4948}{5} = \frac{9896}{10} = 989.6$ $= 4948/5 = 9896/10=989.6$

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What is the average value of the function $f (x) = 2 x^{3} {(1 + x^{2})}^{4}$ $f(x) = 2x^3(1+x^2)^4$ on the interval $[0, 2]$ $[0,2]$ ?

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