I'd like to know how to find the averege values of the following functions
a)f(x)=(x-4)^2 on the interval [2,5]
and
b)f(x)=6sin(x)-3sin(2x) on the interval [0,5]
You find the average of a set of numbers by adding them up and dividing by the number of terms. An integral is essentially a sum of all terms between $\displaystyle a$ and $\displaystyle b$. So, you simply need to find the integral of the function, and divide it by the number of terms. In equation form:
$\displaystyle f(x)_{ave}\vert^b_a = \frac {\int^b_a f(x) dx}{b-a}$
f(averege)=1/(b-a)$f(x)dx...I'm using $ for the integral sign
a)1/3$[2,5](x^2-8x+16)dx
1/3[(x^3/3)-4x^2+16(evaluated from 2 to 5)
1/3[(5^3/3-4(5)^2-16(5))-(2^3/3-4(2)^2+16(2)]
1/3[(125/3-100-80)-(8/3-16+32)]
1/3[(125/3-180)-(8/3 +16)]
1/3(-415/3-56/3)
-471/9-ans