Thread: Averege value of a function

1. Averege value of a function

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]

2. $Ave(f)_{[a,b]} = \frac{{\int\limits_a^b f }}
{{b - a}}$

3. 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 $a$ and $b$. So, you simply need to find the integral of the function, and divide it by the number of terms. In equation form:

$f(x)_{ave}\vert^b_a = \frac {\int^b_a f(x) dx}{b-a}$

thanks, i actually know the formula but i was having trouble with these spicific questions, i might just be making some simple mistake, but i'm not sure what i did wrong

5. Originally Posted by bam017
thanks, i actually know the formula but i was having trouble with these spicific questions, i might just be making some simple mistake, but i'm not sure what i did wrong
Do you know how to find the integral of a function?

6. yes

i'm fimiliar with this type of problem(averege value of a function) and very fimiliar with integration; however, I keep getting the wrong answer on these two problems

7. Originally Posted by bam017
i'm fimiliar with this type of problem(averege value of a function) and very fimiliar with integration; however, I keep getting the wrong answer on these two problems

8. here's what i've got so far

f(averege)=1/(b-a)$f(x)dx...I'm using$ for the integral sign
a)1/3$[2,5](x^2-8x+16)dx 1/3[(x^3/3)-4x^2+16(evaluated from 2 to 5) 1/3[(5^3/3-4(5)^2-16(5))-(2^3/3-4(2)^2+16(2)] 1/3[(125/3-100-80)-(8/3-16+32)] 1/3[(125/3-180)-(8/3 +16)] 1/3(-415/3-56/3) -471/9-ans 9. Originally Posted by bam017 f(averege)=1/(b-a)$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[(125/3-20)-(8/3 +16)]

1/3(39 - 36) = 1
see what a sign error can do ?

btw ... $\int (x-4)^2 dx = \frac{(x-4)^3}{3}$ ... much easier.

10. awesome

of course now it seem so obvious