# Find the average value of the function f over the interval [-1, 2].

f(x) = 1-x^2

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## To find the average value of a function f(x) over an interval [a, b], you can use the formula:

Average value = (1 / (b - a) ) * ∫[a,b] f(x) dx

In this case, the function is f(x) = 1 - x^2, and the interval is [-1, 2]. Let's calculate the average value step by step:

Step 1: Calculate the integral of f(x) over the interval [-1, 2].

∫[-1,2] f(x) dx = ∫[-1,2] (1 - x^2) dx

To integrate the function, we can use the power rule for integration:
∫ x^n dx = (x^(n+1)) / (n+1)

So, applying the power rule, we have:

∫[-1,2] (1 - x^2) dx = [x - (x^3)/3] evaluated from x = -1 to x = 2

= [(2 - (2^3)/3)] - [(-1 - (-1^3)/3)]

= [(2 - 8/3)] - [(-1 + 1/3)]

= [6/3 - 8/3] - [(-3/3 + 1/3)]

= [-2/3] - [-2/3]

= -2/3 + 2/3

= 0

Step 2: Calculate the length of the interval [a, b].

In this case, the interval is [-1, 2], so the length is:

b - a = 2 - (-1) = 3

Step 3: Calculate the average value of the function.

Average value = (1 / (b - a)) * ∫[a,b] f(x) dx

= (1 / 3) * 0

= 0

Therefore, the average value of the function f(x) = 1 - x^2 over the interval [-1, 2] is 0.

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## To find the average value of a function over an interval, we need to find the definite integral of the function over that interval and divide it by the length of the interval.

In this case, we have the function f(x) = 1 - x^2 and the interval [-1, 2].

To find the definite integral of the function, we need to integrate f(x) over the interval [-1, 2]. The integral of f(x) can be found by taking the antiderivative or integrating term by term.

Let's integrate f(x) = 1 - x^2 over the interval [-1, 2]:

∫[from -1 to 2] (1 - x^2) dx

To integrate term by term, we can write the function as:

∫[from -1 to 2] 1 dx - ∫[from -1 to 2] x^2 dx

The integral of 1 with respect to x is x, and the integral of x^2 with respect to x is (x^3)/3.

Using the antiderivative, we get:

= [x] [from -1 to 2] - [(x^3)/3] [from -1 to 2]
= [2 - (-1)] - [(2^3)/3 - ((-1)^3)/3]
=  - [8/3 - (-1/3)]
= 3 - 9/3 + 1/3
= 3 - 8/3
= (9/3) - (8/3)
= 1/3

The definite integral of f(x) = 1 - x^2 over the interval [-1, 2] is 1/3.

Next, we need to find the length of the interval [-1, 2]. The length of an interval can be found by subtracting the smaller endpoint from the larger endpoint:

Length of [-1, 2] = 2 - (-1) = 3

Now, to find the average value of f over the interval, we divide the definite integral by the length of the interval:

Average value of f(x) over [-1, 2] = (1/3) / 3
= 1/9

Therefore, the average value of the function f(x) = 1 - x^2 over the interval [-1, 2] is 1/9.