The Fundamental Theorem of Calculus


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1 The Fundmentl Theorem of Clculus Professor Richrd Blecksmith Dept. of Mthemticl Sciences Northern Illinois University richrd/mth229. The Definite Integrl We define the definite integrl of continuous function f(x) from to b, written f(x)dx to be the re under the curve y = f(x) between two verticl sides x = nd x = b. y = f(x) x = A = f(x)dx x = b then 2. The Fundmentl Theorem of Clculus Prt I If A(x) = the re under the curve y = f(x) from to x, A(x) = x f(t)dt, A (x) = f(x). So in order to solve the re problem, we must find function whose derivtive is f(x). Such functions re just ntiderivtives, which we hve seen before.
2 2 3. Connection to ntiderivtives LetA(x) = x f(t)dt, thereunderthecurvef(t)strtingtndending t x. Note tht we hve written the function s f(t) rther thn s f(x), becuse we re using x s the stopping vlue. Now let F(x) be ny ntiderivtive of f(x). Tht is, F(x) hs the property F (x) = f(x) We hve seen tht A (x) = f(x) So F(x) nd A(x) both hve the sme derivtive f(x). But functions with the sme derivtive differ by constnt. So A(x) = F(x)+C for some constnt C 4. Connection to ntiderivtives II If F(x) is n ntiderivtive of f(x), then A(x) = x f(t)dt = F(x)+C To determine C, plug in x = into the eqution A() = f(t)dt = F()+C Clerly A() = f(t)dt = 0 Why? becuse our intervl [,] is just the single point. So F()+C = 0 or C = F() nd hence A(x) = x f(t)dt = F(x) F()
3 3 Switching letters b for x nd x for t f(x)dx = F(b) F() 5. The Fundmentl Theorem of Clculus f(x)dx = F(b) F() where F(x) is n ntiderivtive of f(x) Clculus ws born when Sir Issc Newton the fig guy? nd Gottfried Wilhelm Leibniz independently discovered this essentil connection between integrls (res) nd derivtives (rtes of chnge). The Fundmentl Theorem of Clculus sys tht in order to find the re A under the curve y = f(x) from to b, first find n ntiderivtive of f(x), tht is, function F(x) whose derivtive is f(x) then the re is A = F(b) F() Compute the definite integrl 6. Esy Exmple 3 2xdx An ntiderivtive of f(x) = 2x is F(x) = x 2 So by the Fundmentl Theorem of Clculus 3 2xdx = F(3) F() = = 8 The re under the line y = 2x for to 3 is 8, s the next picture shows.
4 4 7. Picture 3 2x = 8 y = 2x A 2 =4 (,2) A = 4 A = A +A 2 = 8 (3,6) 8. Some Nottion Insted of writing F(b) F(), we use nottion F(x) b = F(b) F() Evlute the definite integrl 5 2 3x 2 dx An ntiderivtive of f(x) = 3x 2 is F(x) = x 3 So by the Fundmentl Theorem of Clculus 5 2 3x 2 dx = x = = 25 8 = 7 9. Definite vs Indefinite Integrl The integrl sign hs two menings: () without limits nd b, f(x)dx denotes the ntiderivtive of f(x).
5 (2)withlimitsndb, from to b. Type () integrls re clled indefinite integrls. Type (2) integrls re clled definite integrls. f(x)dxdenotesthereunderthecurvey = f(x) 5 0. The +C When employing the Fundmentl Theorem of Clculus f(x)dx = F(b) F() the ntiderivtive F(x) is just the indefinite integrl F(x) = f(x)dx. In this cse, we do not need to write the +C t the end of the integrl becuse it will end up being cncelled. = 8 For exmple, 3 2xdx = ( x 2 +C ) 3 = (3 2 +C) ( 2 +C)= (9+C) (+C)= 9 +C C It is VERY WRONG!!! to write. PC (plus c ) Correctness 3 2xdx = 8+C After ll, the re is the specific number 8. You cnnot dd ny constnt you like to it. On the other hnd, it is lso wrong to forget the +C when writing indefinite integrls, 2xdx = x 2 +C
6 6 2. A Mth Joke Two mle clculus students re sitting in the Junction Resturnt in DeKlb. Mth 229 sure is hrd, sys one of them. Oh, I don t know, sys the other. I bet nybody could lern clculus. Relly? Wht bout our witress? I bet I could tech her some clculus in couple of minutes. Oky, it s bet. The loser buys lunch. So the second fellow wnders off to where the witress (who hppens to be blonde) is filling some drinks t her sttion nd sys to her: 3. End of the Joke When you bring us our check, I ll sk you question. I would like you to nswer: one third x cubed. Cn you do tht? It s worth n extr tip. Sure. The girl giggles nd repets severl times: one third x cubed, one third x cubed, one third x cubed,... When the witress brings the guys their bill, one of them sys, I hve little bet with my friend here. Cn you tell us wht the integrl of x squred is? She replies: one third x cubed. Then, s she leves the tble, the witress sys over her shoulder, plus, of course, n undertermined constnt. 4. Are Problem Find the re under the curve y = x from to 9
7 7 Solution: By the Fundmentl Theorem of Clculus, this re is 9 x /2 dx = 2 3 x3/2 9 = 2 3 [ 9 3 3] = 2 3 (27 ) = Are Problem 2 Find the re under the curve y = sin(x) from 0 to π Solution: In order to use the Fundmentl Theorem of Clculus, we need n ntiderivtive of sin x. Hppily we know function whose derivtive is sinx, nmely the negtive of the cosine cosx. Tht is, sin(x)dx = cos(x)+c π 0 sin(x)dx = cos(x) π 0 = cos(π) ( cos(0)) = + = 2 6. Are Problem 3 Find the re under the curve y = 2x 0 from 3 to 6 6 2x 0dx = x 2 0x 3 = (9 30) = 24+2 = 3 Wit minute. How cn re be negtive? 6 3 The nswer is tht integrls compute the res of regions below the xxis s negtive.
8 8 7. Picture 6 2x 0dx = 3 3 (6,2) y = 2x 0 A = xxis (5,0) A 2 = 4 (3, 4) A = A +A 2 = 3 8. Are between two curves The re between two curves from to b is given by the integrl A = f(x) g(x)dx provided f(x) g(x) on the intervl [,b] Tht is, to find the re between two curves, we lwys subtrct the curve on the bottom from the one on the top. 9. Exmple Find the re y = f(x) = x 2 6x+ nd y = g(x) = x+7. To find where these curve cross, we set f(x) = g(x) x 2 6x+ = x+7 x 2 5x+4 = 0 (x )(x 4) = 0
9 9 x = or x = 4 When x =, y = f() = g() = 6 When x = 4, y = f(4) = g(4) = 3 So the intersection points re (,6) nd (4,3) Now drw the grph to determine which curve is on top. 20. Tle of Two Grphs (,6) y = g(x) = 7 x Region (4,3) y=f(x)=x 2 6x+ 2. Are Between 2 Curves We see from the grph tht g(x) = 7 x lies bove f(x) = x 2 6x+ The re between these two curves is 4 g(x) f(x)dx = 4 (7 x) (x2 6x+)dx = 4 7 x x2 +6x dx = 4 4+5x x2 dx
10 0 = ( 4x+ 5 2 x2 3 x3) 4 = ( ) ( = ) = = = 9 2
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