Let ∫30f(x)dx=6.
(a) What is the average value of f(x) on the interval from x=0 to x=3?
average value = 2
(b) If f(x) is even, find each of the following:∫^3_−3f(x)dx=
the average of f(x) on the interval x=−3 to x=3 =
(c) If f(x) is odd, find each of the following: ∫^3_−3f(x)dx=0/6
the average of f(x) on the interval x=−3 to x=3 = 0
Explanation:
(a) The average value of f(x) on the interval from x=0 to x=3 is given by:
average value = (1/(3-0)) * ∫^3_0f(x)dx
Using the given value of ∫30f(x)dx=6, we can simplify this to:
average value = (1/3) * ∫^3_0f(x)dx = (1/3) * (6/2) = 2
Therefore, the average value of f(x) on the interval from x=0 to x=3 is 2.
(b) If f(x) is even, then we know that f(-x) = f(x) for all x in the domain of f(x). Therefore, we can split the integral from -3 to 3 into two equal intervals from 0 to 3 and from -3 to 0, and use the property of evenness to simplify the integral as follows:
∫^3_−3f(x)dx = ∫^3_0f(x)dx + ∫^0_−3f(x)dx
Since f(x) is even, we have:
∫^3_0f(x)dx = ∫^3_0f(-x)dx
Making the substitution u = -x, du = -dx, the above integral becomes:
∫^0_−3f(x)dx = -∫^0_3f(u)du
Therefore,
∫^3_−3f(x)dx = 2 * ∫^3_0f(x)dx = 2 * (1/3) * ∫^3_0f(x)dx = 2
Therefore, the average of f(x) on the interval x=−3 to x=3 is 2.
(c) If f(x) is odd, then we know that f(-x) = -f(x) for all x in the domain of f(x). Therefore, the integral from -3 to 3 will be equal to zero, since the positive and negative contributions to the integral will cancel out:
∫^3_−3f(x)dx = ∫^3_0f(x)dx + ∫^0_−3f(x)dx
Since f(x) is odd, we have:
∫^3_0f(x)dx = -∫^0_−3f(x)dx
Making the substitution u = -x, du = -dx, the above integral becomes:
∫^0_−3f(x)dx = -∫^3_0f(-u)du = -∫^3_0(-f(u))du = ∫^3_0f(u)du
Therefore,
∫^3_−3f(x)dx = 0
Since the integral is zero, the average of f(x) on the interval x=−3 to x=3 is also zero.