The Fundamental Theorem of CalculusMay 2, 2010The fundamental theorem of calculus has two parts:Theorem (Part I). Let f be a continuous function on [a, b] and define afunction g: [a, b] → R byg(x) :=Zxaf.Then g is differentiable on (a, b), and for every x ∈ (a, b),g0(x) = f(x).At the end points, g has a one-sided derivative, and the same formula holds.That is, the right-handed derivative of g at a is f(a), and the left-handedderivative of f at b is f(b).Proof: This proof is surprisingly easy. It just uses the definition of deriva-tives and the following properties of the integral:1. If f is continuous on [a, b], thenRbaf exists.2. If f is continous on [a, b] and c ∈ [a, b], thenZcaf +Zbcf =Zbaf.3. If m ≤ f ≤ M on [a, b], then(b − a)m ≤Zbaf ≤ (b − a)M.1Let x be a point in (a, b). (We just treat the case of x ∈ (a, b) sincethe endpoints can be treated similarly.) If x ∈ (a, b), we shall show thatg0(x+) = g0(x−) = f(x). Knowing that the two one-sided derivatives existand are equal, we can conclude that the derivative exists and has this value.By definition,g0(x+) = limh→0+g(x + h) − g(x)h.Property (1) assures us that g is well defined provided that h < b − x.Property (2) allows us to simplify the numerator, since it implies thatg(x + h) − g(x) :=Zx+haf −Zxaf =Zx+hxf. (1)This is already great, since we only need to worry about f over the smallinterval [x, x + h]. A picture is helpful here, but I don’t have time to includeone in thise notes. Draw one yourself.Now recall the definition of a limit. We have to show that given any > 0, there is a δ > 0 such thatg(x + h) − g(x)h− f(x)< . (2)whenever 0 < h < δ. This is where we use the continuity of f at x. We knowfrom this that there is a δ such that |f (x0) − f(x)| < whenever |x0− x| < δ.This means thatf(x) − < f(x0) < f(x) + .for all such x0. We use this same δ our criterion for the limit in equation (2).Let us verify that this works. Suppose that 0 < h < δ. Then on the interval[x, x + h], we know that f is between f(x) − and f(x) + . By property (3)of integrals, it follows that(f(x) − )h ≤Zx+hxf ≤ (f(x) + )h.Since h > 0, we can divide both sides by h to conclude thatf(x) − ≤ 1/hZx+hxf ≤ f(x) + , i.e.,f(x) − ≤g(x + h) − g(x)h≤ f(x) + 2This is exactly what we needed.The left handed derivatives are done in essentially the same way.Theorem (Part II). Let f be a continuous function on [a, b]. Supposethat F is continuous on [a, b] and that F0= f on (a, b). ThenZbaf = F(b) − F (a).Proof: Consider the function g in the previous theorem. Since g is differ-entiable on [a, b] it is continuous there (including at the end points, wherethe one-sided deritaives exist). We also know that g and F are differentiableon (a, b), and that there derivatives are equal. Recall that we had (as aconsequence of the mean value theorem for derivatives) that F and g differby a constant. That is, there is a number C such that g(x) = F (x) for allx ∈ [a, b]. ThenF (b) − F (a) = (g(b) + C) − (g(a) + C) = g(b) − g(a) =Zbaf −Zaaf
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