Find the average value of the function f over the interval [-1, 2].
f(x)=1-x^2
=1/(2-(-1) ∫((1/2)x^2 - x + 3) dx from x = -1 to 2
=1/3 [ (1x-x^3/3 ] from x = -1 to 2
=1/3[2-8/3)-(1-1/3)]
1/3[-2/3-2/3] =0
It still says it is wrong. =/
I get zero.
∫[-1,2] 1-x^2 dx
= x - x^3/3 [-1,2]
= (2 - 8/3) - (-1 + 1/3)
= 0
yeah, I know, we have to divide by 3 ...
To find the average value of a function over an interval, you need to:
1. Evaluate the function at all points within the interval.
2. Calculate the sum of the function values.
3. Divide the sum by the length of the interval.
Let's work through these steps to find the average value of the function f(x) = 1 - x^2 over the interval [-1, 2].
Step 1: Evaluate the function at all points within the interval:
- Plug in the lower bound, -1, into the function: f(-1) = 1 - (-1)^2 = 1 - 1 = 0.
- Plug in the upper bound, 2, into the function: f(2) = 1 - 2^2 = 1 - 4 = -3.
Step 2: Calculate the sum of the function values:
- Add the function values at the lower and upper bounds: 0 + (-3) = -3.
Step 3: Divide the sum by the length of the interval:
- The length of the interval [-1, 2] is 2 - (-1) = 3.
- Divide the sum (-3) by the length of the interval (3): -3/3 = -1.
Therefore, the average value of the function f(x) = 1 - x^2 over the interval [-1, 2] is -1.