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 =
(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=
the average of f(x) on the interval x=−3 to x=3=
(a) The average value of f(x) on the interval from x=0 to x=3 is given by:
average value = 1/(3-0) * ∫0^3 f(x) dx = 1/3 * 6 = 2
(b) If f(x) is even, then we know that f(-x) = f(x) for all x. Therefore, we can split the integral from -3 to 3 into two parts, and use the fact that f(x) is even to simplify:
∫3−3f(x)dx = 2 * ∫0^3 f(x) dx
(the factor of 2 comes from the fact that the integral is over a symmetric interval)
We already know that ∫0^3 f(x)dx = 6/3 = 2, so substituting in:
∫3−3f(x)dx = 2 * 2 = 4
To find the average value of f(x) on the interval x = -3 to x = 3, we use the same formula as before:
average value = 1/(3-(-3)) * ∫-3^3 f(x) dx = 1/6 * 4 = 2/3
(c) If f(x) is odd, then we know that f(-x) = -f(x) for all x. Therefore, when we integrate over a symmetric interval, the integral will be zero:
∫3−3f(x)dx = 0
To find the average value of f(x) on the interval x = -3 to x = 3, we again use the formula:
average value = 1/(3-(-3)) * ∫-3^3 f(x) dx = 0/6 = 0