Interpreting Integrals and the Fundamental Theorem


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1 Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of some quntity, then f (t) dt is the totl chnge in the quntity over the intervl from t = to t =. For emple: Velocity v(t) is the rte of chnge of the position of n oject, so v(t) dt 5 If f(t) is the rte of chnge of CO concentrtion in pond wter, then f (t) dt 0 4 In ech cse, we think of multiplying rte of chnge times n intervl (usully time intervl) to get the totl chnge. The units help revel the kind of quntity tht the integrl represents. For ech term in Riemnn sum, we multiply function vlue times t or : If the units of velocity function v(t) re in feet per second nd the time t is 5 mesured in seconds, then v(t) dt must e If f(t) is the rte of chnge of CO concentrtion in pond wter, with units mmol/l per hour, then 0 4 f (t) dt In generl, we see tht we will lwys get Units on f (t) dt = Let s get some prctice. In the following, rememer tht: ) Suppose v(t) gives the velocity (in hundreds of inches per minute) of n oject t time 7 t (given in minutes). Then wht does v(t)dt =6 men in words? Wht units is the si in? 4
2 ) Wter is leking from tnk t rte of R(t) gllons per minute. Set up n integrl to find the mount of wter tht leks out in the first seven minutes. ) If f(z) represents the numer of flus per zork when the level of zorks is z, wht does f (z)dz represent? Wht re the units? 4) If f(z) insted represents the numer of flus when the level of zorks is z, wht re the units on f (z) dz? 5) Following lek of rdioctive mteril in smll town, we estimte there is rdition level t hours fter the lek of R(t) =.e.0t millirems/hour. ) Suppose it hs een decided tht n eposure of.05 millirems/hour is sfe. How long will it tke efore we re t sfe level gin fter the lek? ) Set up n integrl to evlute the totl mount of rdition relesed until the rdition rte is ck to n cceptle level. 6) Suppose we re producing Hokie ook covers for sle in the ook store, nd profit function P() (in dollrs) from selling hundred ook covers hs derivtive P () = 6. ) Where is profit incresing nd where is it decresing? ) How mny covers should we mke to get the mimum profit? c) Set up n integrl to represent the totl profit from selling tht mny ook covers. 5
3 Fundmentl Theorem Version I: Interpreting Integrls We hve rrived t the cru of this course. The fundmentl theorem of clculus hs two prts, which will mke the connection etween derivtives nd integrls, thus linking the two min topics we hve studied in 06 nd 05. Fundmentl Theorem of Clculus I: Let s try putting wht the fundmentl theorem sys in words, nd see why it might e true. Since F'() = f(), wht we re relly integrting is the rte of chnge of F. On the other hnd, F() F() is simply the difference etween the originl nd the finl F vlue. Thus, F() F() is ctully the totl chnge in F etween nd. Thus, in words, the fundmentl theorem sys: But we lredy knew this! One impliction of the fundmentl theorem is tht if we cn find function F whose derivtive is f, then we cn evlute f () d simply y evluting F t the two endpoints of the intervl nd sutrcting. Let s look some emples of estimting totl chnge from the rte of chnge: Emple: Let s revisit the emple of selling ook covers from erlier. Suppose we know tht the derivtive of the profit function from selling hundred ook covers is P () =6. Suppose lso tht we know we will mke $6 from selling 00 ook covers. Wht will e our profit from selling 600 ook covers? In this cse, we know tht P() = 6. We wnt to find the dditionl profit from selling etween 00 nd 600 covers, i.e., where goes from to 6. According to the fundmentl theorem, So the dditionl profit cn e computed y 6 d, which we cn esily clculte using Simpson s rule with n =, which mkes = : 6 6
4 Thus, the dditionl profit (the chnge in profit) is, for totl profit of from selling 600 ook covers. Emple: Suppose the rte of chnge of the fish popultion in given pond is given y the grph elow, where the is gives time in months, nd the yis gives hundreds of fish per month. Is the finl popultion lrger or smller thn the initil popultion? Determine where the fish popultion is highest nd lowest during the month cycle shown. Highest: Lowest: Finl popultion Hundred Fish Per Month Month Emple: Let s clculte d using the fundmentl theorem. First we need function F() with derivtive F'() =. (Such function is clled n ntiderivtive of.) One such function is F() = Now we just need to clculte F() F() for this integrl. In this cse of course, = nd =. So d = In this cse, it s esy to check, since we could hve used simple geometry or the trpezoidl rule to find the ect vlue of this liner function: 7
5 4 Use Trp Rule use re So it seems the fundmentl theorem is correct! Of course, ll the things we knew out integrls re still in effect, so we cn use ides such s res to interpret totl chnges s well. It my e chllenging to come up with such n F. We ll study how to do this it lter; for now, it will e it difficult to use version one to evlute integrls. Let s do nother emple for prctice: Emple: If F() = 9, nd, F () =, estimte F(7). We know tht We now need to go on rief ecursion efore we cn introduce second version of the fundmentl theorem. 8
6 Integrl Defined Functions Suppose we hve continuous function f. We cn define new function F sed on f y picking point to strt integrting from nd letting Note the following: The point is constnt. F() = f (t)dt The vrile is the vrile for F nd descries how fr to integrte from. The vrile t is clled dummy vrile; it is used to hold the plce of vrile inside the integrl when we integrte. We cn think of F() s eing the re under the curve f from up to, if we rememer to interpret res elow the is ppropritely. Emple: Suppose f () = +. Let = nd let F() = f (t) dt. Estimte F() nd F(). We know tht F() = + t dt. (We switched the vrile to t inside the integrl ecuse we re using for the vrile which descries how fr to integrte.) Thus, F() = + t dt. We will pproimte this with Simpson s rule nd n = just to hve quick, rough estimte: Similrly, F() = + t dt. We will estimte this with Simpson s rule with the sme t (nd thus n = 4 suintervls): Therefore, we estimte F() is out, nd F() is out. Wht do you think F() is? 9
7 Fundmentl Theorem II: Derivtives of Integrls Now tht we know how to define functions in terms of integrls, we wnt to know how to tke the derivtive of such function. The second hlf of the Fundmentl Theorem nswers this question. Theorem (Fundmentl Theorem of Clculus II): Note crefully wht this theorem sys! It sys tht if we hve n integrl defined function nd we wnt to find the derivtive, ll we do is tke the integrnd out nd stick the vrile into it. (Fundmentl Theorem II lso sys tht F() is one ntiderivtive of f().) Emple: Suppose we define F() = cos(t ) dt. Then F () =.5 Emple: Suppose F() = t + t + dt. Then F () = 7 Emple: Suppose f is function given y the grph elow: Use the Fundmentl Theorem to identify where F() = where it is decresing. f (t) dt is incresing nd 0 0
8 Summry Tody, we hve Mde eplicit our ide out integrls giving the totl chnge sed on rte of chnge. Used units to help interpret the mening of definite integrl. Stted version one of the Fundmentl Theorem of Clculus, which essentilly sys symoliclly tht the integrl of rte of chnge is the totl chnge: F () d =F() F () Used the integrl of derivtive ( rte of chnge) to estimte new vlues of function, sed on strting vlue. Discussed the ide of functions defined in terms of integrls, nd used the Fundmentl Theorem of Clculus version two to find derivtives of such functions.
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