Find the average value over the given interval.

y=3x^5; [-3,3]

I can't seem to get the answer, 0.
1/(3+3) [integral -3 to 3 (3x^5)dx]
and I get 273/2

I took the antiderivative incorrectly. I figured out my mistake.

To find the average value of a function over a given interval, you need to evaluate the definite integral of the function over that interval and then divide it by the width of the interval.

In this case, you have the function y = 3x^5 and the interval [-3, 3].

To find the average value, you need to evaluate the definite integral of the function y = 3x^5 over the interval [-3, 3].

The integral of the function y = 3x^5 can be found by using the power rule for integration:

∫(3x^5)dx = (3/6)x^6 + C = x^6/2 + C

Then, you need to evaluate the integral at the upper and lower limits of the interval:

∫(3x^5)dx evaluated from -3 to 3 is [(3^6/2 + C) - (-3^6/2 + C)]

Simplifying further, you get:

[(3^6/2 + C) - (-3^6/2 + C)] = [(729/2 + C) - (729/2 + C)] = 0

However, this is not the final answer. To find the average value, you need to divide this integral by the width of the interval, which is 3 - (-3) = 6. So, the average value is:

0 / 6 = 0

Therefore, the average value of the function y = 3x^5 over the interval [-3, 3] is 0.

It seems like there might have been an error in your calculations or integration steps that resulted in a different answer.