The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at the same point in that interval. The theorem guarantees that if f(x) is continuous, a point c exists in an interval [a, b] such that the value of the function at c is equal to the average value of f(x) over [a, b]. We state this theorem mathematically with the help of the formula for the average value of a function that we presented at the end of the preceding section.
If f(x) is continuous over an interval [a, b], then there is at least one point c ∈ [a, b] such that
Using this theorem, we can find the value of x in the interval [a, b], where f(x) attains its average value.
Example 1 :
Find the average value of the function f(x) = 2x + 6 over the interval [0, 7] and find c such that f(c) equals the average value of the function over [0, 7].
Solution :
Average value of the function :
Set the average value equal to f(c) and solve for c.
f(c) = 13
2c + 6 = 13
2c = 7
c = 3.5
Example 2 :
Find the average value of the function f(x) = ˣ⁄₂ over the interval [0, 6] and find c such that f(c) equals the average value of the function over [0, 6].
Solution :
Average value of the function :
Set the average value equal to f(c) and solve for c.
f(c) = 1.5
ᶜ⁄₂ = 1.5
c = 3
Example 3 :
Find the average value of the function f(x) = x^{2} over the interval [0, 3] and find c such that f(c) equals the average value of the function over [0, 3].
Solution :
Average value of the function :
Set the average value equal to f(c) and solve for c.
f(c) = 3
c^{2} = 3
c = ±√3
c ≈ -1.732 or 1.732
-1.732 ∉ [0, 3] and 1.732 ∈ [0, 3]
Thus,
c ≈ 1.732
Example 4 :
Find the value of x, where the function f(x) = 2x + 3 attains its average value over the interval [0, 5].
Solution :
Let c be the value of x, where the function f(x) attains its average value over the interval [0, 5].
2c + 3 = 8
2c = 5
c = 2.5 ∈ [0, 5]
The given function attains its average value at x = 2.5 over the interval [0, 5].
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