4XL !"#$%&'()B–ú LkBhc;4Two Fundamental Theorems about the Definite Integral*kBhcàÙThese lecture notes develop the theorem Stewart calls The Fundamental Theorem of Calculus in section 5.3. The approach I use is slightly different than that used by Stewart, but is based on the same fundamental ideas.)kBhcThe definite integralkBhc!Recall that the expressionetnIGkBhc f(x)kBhcxkBhcbkBhca kBhc*#is called the definite integral of kBhc f(x)kBhc over the interval kBhc [a,b]kBhc5. and stands for the area underneath the curve kBhcy = f(x)kBhc over the interval kBhc [a,b]kBhc5. (with the understanding that areas above the kBhcxkBhc]V-axis are considered positive and the areas beneath the axis are considered negative).!kBhcÿIn today's lecture I am going to prove an important connection between the definite integral and the derivative and use that connection to compute the definite integral. The result that I am eventually going to prove sits at the end of a chain of earlier kBhc,%definitions and intermediate results.kBhc6/Some important facts about continuous functions"kBhcüThe first intermediate result we are going to have to prove along the way depends on some definitions and theorems concerning continuous functions. Here are those definitions and theorems.kBhc#The definition of continuitykBhc A function kBhc f(x)kBhc! is continuous at a point kBhc x = akBhc if the following holdkBhc f(a)kBhc existstmiLkBhc xÏakBhc  fGrapkBhcxkBhc existstmiLkBhc xÏakBhc  fGrapkBhcxkBhc  = fGrapkBhcakBhc A function kBhc f(x)kBhc% is continuous in an interval kBhc [a,b]kBhc<5 if it is continuous at every point in that interval. kBhc The extreme value theorem kBhc Let kBhc f(x)kBhc0) be a continuous function in an interval kBhc [a,b]kBhc. The exist numbers kBhcmkBhc  and kBhcMkBhc such that kBhc m ² f(x) ² M kBhcfor all kBhcxkBhc  in kBhc [a,b]kBhc(!. Furthermore, there are numbers kBhcckBhc  and kBhcdkBhc  in kBhc [a,b]kBhc such that kBhcf(c) = mkBhc  and kBhcf(d) = MkBhc. kBhc%The intermediate value theoremkBhc Let kBhc f(x)kBhc0) be a continuous function in an interval kBhc [a,b]kBhc , and let kBhcNkBhc be any number between kBhc f(a)kBhc  and kBhc f(b)kBhc . Then there is a number kBhcckBhc  in kBhc [a,b]kBhc such that kBhcf(c) = NkBhc.kBhc70A preliminary result about the definite integral kBhcTheoremkBhc  Let kBhc f(x)kBhc1* be a continuous function on the interval kBhc [a,b]kBhc. Then there exists a kBhcckBhc  in kBhc [a,b]kBhc for whichkBhcfGrapkBhcckBhc GrapkBhc b - akBhc  = etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhcE>This theorem essentially says that if you take the area under kBhc f(x)kBhc over the interval kBhc [a,b]kBhcsl and Ôflatten it outÕ, you get a rectangle whose height is given by the value of the function at some point kBhcckBhc in the interval.kBhc ProofkBhc Because kBhc f(x)kBhc is continuous on kBhc [a,b]kBhcle, the extreme value theorem tells us it must have a maximum and a minimum somewhere in that interval.kBhc m ² fGrapkBhcxkBhc  ² MkBhc-&This leads to the inequality for areas kBhc m GrapkBhc b - akBhc = etnIDkBhcmkBhcxkBhcbkBhcakBhc  ² etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc  ² etnIDkBhcMkBhcxkBhcbkBhcakBhc  = M GrapkBhc b - akBhc orkBhc m ² carF|etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc b - akBhc  ² M kBhc Since kBhc f(x)kBhc_X is continuous, by the Intermediate Value Theorem it takes every possible value between kBhcmkBhc  and kBhcMkBhc4-. In particular, there is at least one place kBhcckBhc at which the function kBhc f(x)kBhc has a value equal to kBhc  fGrapkBhcckBhc  = carF|etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc b - akBhc!Multiplying both sides by kBhc b - akBhc proves the result.kBhc92The first fundamental theorem of integral calculus#kBhcŒ…We are now in a position to prove our first major result about the definite integral. The result concerns the so-called area function$kBhcF(x) = etnI\kBhcfGrapkBhctkBhctkBhcxkBhca&kBhc*#and its derivative with respect to kBhcxkBhcG@. The only way to compute this derivative is via the definition:'kBhcFopxEkBhcákBhc (x) = tmiLkBhc h Ï 0kBhc carF0kBhc F(x+h) - F(x)kBhch(kBhc = tmiLkBhc hÏ0carFèetnI^kBhcfGrapkBhctkBhctkBhc x+hkBhcakBhc  - etnI\kBhcfGrapkBhctkBhctkBhcxkBhcakBhch)kBhcc\Since the numerator in the fraction is the difference of two areas, we can rewrite the limit*kBhc = tmiLkBhc hÏ0kBhc carFzetnI^kBhcfGrapkBhctkBhctkBhc x+hkBhcxkBhch+kBhc)"Recall the theorem we just proved.,kBhc  fGrapkBhcckBhc  = carF|etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc b - a-kBhc3,If we apply that theorem to the fraction in feR *(1)kBhc  with.kBhc a = x/kBhcb = x+h0kBhc b - a = h1kBhc we get2tmiLkBhc hÏ0kBhc carFzetnI^kBhcfGrapkBhctkBhctkBhc x+hkBhcxkBhchkBhc  = tmiLkBhc hÏ0kBhc  fGrapkBhcc3 kBhc where kBhcckBhc% is somewhere in the interval kBhc[x,x+h]kBhc. In the limit as kBhchkBhc goes to 0, kBhcckBhc gets squeezed down to kBhcxkBhc . Because kBhc f(x)kBhc" is continuous we have that4 tmiLkBhc hÏ0kBhc  fGrapkBhcckBhc  = tmiLkBhc c Ï xkBhc  fGrapkBhcckBhc  = fGrapkBhcx5kBhcThe bottom line is thatkBhcFopxEkBhcákBhc (x) = tmiLkBhc h Ï 0kBhc carF0kBhc F(x+h) - F(x)kBhchkBhc  = fGrapkBhcx6kBhc:3The second fundamental theorem of integral calculusLkBhcjcWe are now in a position to prove the final and most important theorem in this sequence of results.8kBhcTheoremkBhc  Let kBhc f(x)kBhc1* be a continuous function on the interval kBhc [a,b]kBhc . Let kBhc G(x)kBhc.' be any function with the property that9kBhcGopxEkBhcákBhc(x) = fGrapkBhcx:kBhc Then;etnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc = G(b) - G(a)<kBhc ProofkBhc0) As above, we introduce the area function=kBhcF(x) = etnI\kBhcfGrapkBhctkBhctkBhcxkBhca>kBhc(!By the first fundamental theorem,?kBhcFopxEkBhcákBhc (x) = f(x)@kBhc Since kBhc F(x)kBhc  and kBhc G(x)kBhc_X have the same derivative, they must differ by a constant (Corollary 4.2.7 on page 294).AkBhcF(x) = G(x) + CBkBhc3,What is that constant? Evaluating equations feR =(2)kBhc  and feR A(3)kBhc  at kBhc x = akBhc gives usCkBhcF(a) = G(a) + C = etnI\kBhcfGrapkBhctkBhctkBhcakBhcakBhc  = 0DkBhcwhich leads toEkBhc G(a) + C = 0FkBhc C = -G(a)GkBhc.'Similarly, evaluating the equations at kBhc x = bkBhc  givesHkBhcF(b) = G(b) + C = etnI\kBhcfGrapkBhctkBhctkBhcbkBhcaIkBhc orJetnI\kBhcfGrapkBhctkBhctkBhcbkBhcakBhc  = G(b) + C = G(b) - G(a)KkBhcand we are done.