Find the average value of the function f over the interval [-1, 2].
f(x) = 1-x^2
To find the average value of a function f(x) over an interval [a, b], you need to calculate the definite integral of the function over that interval and divide it by the width of the interval (b - a).
Let's find the average value of the function f(x) = 1 - x^2 over the interval [-1, 2].
Step 1: Calculate the definite integral of f(x) over the interval [-1, 2].
∫[a, b] f(x) dx = ∫[-1, 2] (1 - x^2) dx
To integrate 1 - x^2, we can use the power rule of integration:
∫(x^n) dx = (x^(n+1))/(n+1) + C
∫(1 - x^2) dx = ∫1 dx - ∫(x^2) dx
= x - (x^3/3) + C
Evaluating the definite integral:
∫[-1, 2] (1 - x^2) dx = (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 width of the interval.
b - a = 2 - (-1)
= 2 + 1
= 3
Step 3: Divide the result from Step 1 by the result from Step 2.
Average value = 0 / 3
= 0
Therefore, the average value of the function f(x) = 1 - x^2 over the interval [-1, 2] is 0.