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Chapter 5 | Integration
Figure 5.27 Over the interval [0, 3], the function f ( x ) = x 2 takes on its average value at c = 3.
3 ⎛
5.15
⎝ 2 x 2 −1 ⎞
Given ∫
⎠ dx =15, find c such that f ( c ) equals the average value of f ( x ) =2 x 2 −1 over
0
[0, 3].
Fundamental Theorem of Calculus Part 1: Integrals and Antiderivatives
As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of Calculus, Part 1 , is stated here. Part 1 establishes the relationship between differentiation and integration.
Theorem 5.4: Fundamental Theorem of Calculus, Part 1 If f ( x ) is continuous over an interval ⎡ ⎣ a , b ⎤
⎦ , and the function F ( x ) is defined by
x
(5.16)
F ( x ) = ∫
f ( t ) dt ,
a
then F ′( x ) = f ( x ) over ⎡
⎤ ⎦ .
⎣ a , b
Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that we have defined a function, F ( x ), as the definite integral of another function, f ( t ), from the point a to the point x . At first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like it’s a function. The key here is to notice that for any particular value of x , the definite integral is a number. So the function F ( x ) returns a number (the value of the definite integral) for each value of x .
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