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

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Answer 1 · 2018-01-06T10:22:34

${f (x)}_{a v g} = \frac{8}{3}$ $f(x)_(avg)=8/3$

OR

${f (x)}_{a v g} = 2.66$ $f(x)_(avg)=2.66$

The average value of a function $= \frac{\int_{a}^{b} f (x) d x}{b - a}$ $=(int_a^bf(x)dx)/(b-a)$

Here $f (x) = 4 x - x^{2}$ $f(x)=4x-x^2$ and $b = 4$ $b=4$ , $a = 0$ $a=0$

So,

${f (x)}_{a v g} = \frac{\int_{0}^{4} (4 x - x^{2}) d x}{4 - 0}$ $f(x)_(avg)=(int_0^4(4x-x^2)dx)/(4-0)$

${f (x)}_{a v g} = \frac{{[\frac{4 x^{2}}{2} - \frac{x^{3}}{3}]}_{0}^{4}}{4}$ $f(x)_(avg)=[(4x^2)/2-x^3/3]_0^4/4$

${f (x)}_{a v g} = \frac{{[2 x^{2} - \frac{x^{3}}{3}]}_{0}^{2}}{4}$ $f(x)_(avg)=[2x^2-x^3/3]_0^4/4$

${f (x)}_{a v g} = \frac{(2 \times 4^{2} - \frac{4^{3}}{3}) - (2 \times 0^{2} - \frac{0^{3}}{3})}{4}$ $f(x)_(avg)=((2xx4^2-4^3/3)-(2xx0^2-0^3/3))/4$

${f (x)}_{a v g} = \frac{\frac{32}{3}}{4}$ $f(x)_(avg)=(32/3)/4$

${f (x)}_{a v g} = \frac{8}{3}$ $f(x)_(avg)=8/3$

${f (x)}_{a v g} = 2.66$ $f(x)_(avg)=2.66$