1. ## Finding Average Value

1) Find the average value of the function over the indicated interval:
x^2-sin(x) on [-2,4].

I am not sure how to go about solving this problem. Any help on it will be much appreciated!

-M

2. Originally Posted by blurain
1) Find the average value of the function over the indicated interval:
x^2-sin(x) on [-2,4].

I am not sure how to go about solving this problem. Any help on it will be much appreciated!

-M
The average value of a function $\displaystyle f(x)$ on an interval $\displaystyle [a,b]$ is given by:

$\displaystyle \mbox{Average Value } = \frac 1{b - a} \int_a^b f(x)~dx$

3. Thank you for replying. However, I already know the formula to find the average value; I just do not understand the algebraic part of the problem.

-M

4. Originally Posted by blurain
Thank you for replying. However, I already know the formula to find the average value; I just do not understand the algebraic part of the problem.

-M
by "algebraic" do you mean finding the integral $\displaystyle \frac 16 \int_{-2}^4 \left( x^2 - \sin x \right)~dx$ ?

5. Ah, yes, that was what I meant.

6. Originally Posted by blurain
Ah, yes, that was what I meant.
in general, $\displaystyle \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $\displaystyle n \ne -1$

if $\displaystyle n = -1$, $\displaystyle \int x^{-1}~dx = \ln x + C$

$\displaystyle \int \sin x ~dx = - \cos x + C$

now can you continue?

you do know how to plug in the limits of integration according to the fundamental theorem of calculus, right?

7. Would that mean that x^2-sin(x) is 1/3(x^3)+cos(x)?

8. Originally Posted by blurain
Would that mean that x^2-sin(x) is 1/3(x^3)+cos(x)?
yes, that's the integral. now plug in the limits. and don't forget that you have a 1/6 in front