What is the average value of a function f(x) x^2 - 2x + 5f(x)x2−2x+5 on the interval [2,6]?

1 Answer
ali ergin
Apr 21, 2016

k=172/12k=17212

Explanation:

"let's recall that the average value of a function for an interval"let's recall that the average value of a function for an interval
" of (a,b) is given by formula:" of (a,b) is given by formula:
k=1/(b-a)int_a^b f(x) d xk=1b−a∫baf(x)dx
where;where;
k:"average value"k:average value
k=1/(6-2)int_2^6(x^2-2x+5)d xk=16−2∫62(x2−2x+5)dx

k=1/4(|x^3/3-2x^2/2+5x |_2^6)k=14(∣∣∣x33−2x22+5x∣∣∣62)

k=1/4(|x^3/3-x^2+5x|_2^6)k=14(∣∣∣x33−x2+5x∣∣∣62)

k=1/4[(6^3/3-6^2+5*6)-(2^3/3-2^2+5*2)]k=14[(633−62+5⋅6)−(233−22+5⋅2)]

k=1/4[(216/3-36+30)-(8/3-4+10)]k=14[(2163−36+30)−(83−4+10)]

k=1/4[(216/3-6)-(8/3+6)]k=14[(2163−6)−(83+6)]

k=1/4[(216-18)/3-(8+18)/3]k=14[216−183−8+183]

k=1/4[198/3-26/3]k=14[1983−263]

k=1/4[172/3]k=14[1723]

k=172/12k=17212

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