Consider the given function and the given interval.
f(x) = (x − 3)^2, [2, 5]
(a) Find the average value fave of f on the given interval.
fave =
(b) Find c such that fave = f(c).
c = (smaller value)
c= (larger value)
recall that the average value of f(x) on [a,b] is
(∫[a,b] f(x) dx)/(b-a)
now just plug and chug.
To find the average value (fave) of the function f(x) on the given interval [2, 5], you need to perform the following steps:
(a) Calculate the definite integral of f(x) over the interval [2, 5].
The average value of a function on an interval is given by:
fave = (1 / (b - a)) * ∫[a to b] f(x) dx,
where ∫[a to b] f(x) dx represents the definite integral of f(x) over the interval [a, b], and (b - a) is the length of the interval.
In this case, the interval is [2, 5]:
a = 2 (lower limit)
b = 5 (upper limit)
Calculating the definite integral of f(x) = (x - 3)^2 on the interval [2, 5]:
∫[2 to 5] (x - 3)^2 dx = [((x - 3)^3 / 3)] evaluated from 2 to 5
= ((5 - 3)^3 / 3) - ((2 - 3)^3 / 3)
= (2^3 / 3) - (-1^3 / 3)
= (8/3) - (-1/3)
= (8 + 1) / 3
= 9 / 3
= 3
Now, divide the result by the length of the interval to find the average value (fave):
fave = (1 / (5 - 2)) * ∫[2 to 5] (x - 3)^2 dx
= (1 / 3) * 3
= 1
Therefore, the average value of f(x) on the interval [2, 5] is fave = 1.
(b) To find c such that fave = f(c), set the average value fave equal to the function f(x), and solve for c.
Since fave = 1 (as found in part a), we need to solve the equation f(c) = 1 for c.
In this case, the function f(x) is given by f(x) = (x - 3)^2.
Setting f(c) equal to 1:
(x - 3)^2 = 1
Taking the square root of both sides:
x - 3 = ±1
Solving for x:
Case 1: x - 3 = 1
x = 4
Case 2: x - 3 = -1
x = 2
Therefore, the two values of c that satisfy fave = f(c) = 1 are:
c = 2 (smaller value)
c = 4 (larger value)
Hence, c = 2 and c = 4 are the values for which fave = f(c) = 1.
(a) To find the average value fave of f on the interval [2, 5], we need to evaluate the integral of f(x) over the interval and divide it by the length of the interval.
First, let's find the integral of f(x) over the interval [2, 5]:
∫[2, 5] (x − 3)^2 dx
To evaluate this integral, we can expand the square and then integrate:
∫[2, 5] (x^2 - 6x + 9) dx
= ∫[2, 5] x^2 dx - ∫[2, 5] 6x dx + ∫[2, 5] 9 dx
= (1/3)x^3 - 3x^2 + 9x ∣[2, 5]
= [(1/3)(5)^3 - 3(5)^2 + 9(5)] - [(1/3)(2)^3 - 3(2)^2 + 9(2)]
= (125/3 - 75 + 45) - (8/3 - 12 + 18)
= (95/3) - (14/3)
= 81/3
= 27
Next, let's find the length of the interval [2, 5]:
length = 5 - 2 = 3
Finally, the average value fave of f on the interval [2, 5] is:
fave = (1/length) * ∫[2, 5] (x − 3)^2 dx
= (1/3) * 27
= 9
Therefore, the average value fave of f on the interval [2, 5] is 9.
(b) To find c such that fave = f(c), we can set f(c) equal to the average value fave and solve for c:
f(c) = fave
(c − 3)^2 = 9
Taking the square root of both sides, we get:
c − 3 = ±√9
c − 3 = ±3
Solving for c, we have two cases:
Case 1: c - 3 = 3
c = 3 + 3
c = 6
Case 2: c - 3 = -3
c = 3 - 3
c = 0
So, the values of c such that fave = f(c) are c = 0 (smaller value) and c = 6 (larger value).