Setions 5.3: Antierivtives n the Funmentl Theorem of Clulus Theory: Definition. Assume tht y = f(x) is ontinuous funtion on n intervl I. We ll funtion F (x), x I, to be n ntierivtive of f(x) if F (x) = f(x) for x I. Remrk. We wrote F (x), x I, to be n ntierivtive of f(x) n not the ntierivtive beuse ntierivtives re not unique: x 3 x3 3 s well 3 + 7 re ntierivtives of the funtion y = x2. More generlly, if F (x) is n ntierivtive of funtion f(x) for ny then for every onstnt the funtion G(x) = F (x) + is lso n ntierivtive of f(x). As the next theorem shows ll ntierivtives re of tht form. Theorem. Assume tht y = f(x) is ontinuous funtion on n intervl I. Let F (x), x I, be n ntierivtive of f(x). Then for every onstnt F (x) + is lso n ntierivtive of f(x). Conversely, if G(x), x I, is ny ntierivtive of f(x), then there is onstnt, so tht G(x) = F (x) +, for x I. Proof. Let F (x) be n ntierivtive of f(x) on I n onstnt. Then ( ) F (x) + = F (x) = f(x) + 0 = f(x). Conversely let G(x) be ny ntierivtive on I, then ( ) G(x) F (x) = G(x) F (x) = f(x) f(x) = 0. Now, by Theorem 5 of Setion 5.3, this implies tht F (x) G(x) is onstnt on I, n thus F (x) = G(x) +, for x I.
2 The Funmentl Theorem of Clulus: As the nme lrey suggests, this is n immensely importnt theorem. It onnets the omputtion of efinite integrls n the omputtion of ntierivtives. Theorem 2. Assume tht y = f(x) is ontinuous funtion on n intervl I = [, b]. Prt I of the Funmentl Theorem of Clulus: Define G(x) = x Then G(x) is n ntierivtive of f. f(t)t, for x (, b). Prt II of the Funmentl Theorem of Clulus: If F (x), x I, is ny ntierivtive of f(x). Then (Problem 3) Exmple. Compute b f(x) = F (b) F (). 5 2 x 2. So oring to prt II of the FTC. We nee to fin n nti erivtive of f(x) = x 2. Now F (x) = x3 3 is n ntierivtive. Thus 5 x 2 = x3 5 3 = 23 2 3 53 3. 2
Proof. Prt I of the FTC: We nee to prove for < < b tht G G( + h) G(x) () = lim = f(). h 0 h Let us first ssume tht h > 0 n tht f() is non negtive. Then ( ( ) +h ) G( + h) G() = f(x) f(x) = +h f(x)t. h h h In our piture below +h f(t)t represents the re of the region boune by the the vertil lines x = n x + h, bove the x-xis, n the grph of y = f(x), t + h. This region ontins smll retngle whose sies re h n f(x min,h ) where f(x min,h ) is the miniml funtion vlue on [, + h] (whih in our piture is f(x), but oes not nee to be). The region is ontine ontine in lrge retngle whose sies re h n f(x min,h ) where f(x mx,h ) is the mximl funtion vlue on [x, x+h] (whih in the piture, is f(+h)). Thus n, by iviing by h h f(x min,h ) f(x min,h ) h +h +h f(x)h f(x mx,h ), f(x) f(x mx,h ). Now lim h 0 + x min,h = n lim h 0 + x mx,h =, n thus, sine f(x) is ontinuous y lim f(x min,h) = f() n h 0 + lim f(x mx,h) = f() h 0 + 3 x x+h b x
4 We therefore eue from the Squeeze Theorem tht: G( + h) G(x) lim = lim h 0 + h h 0 + h We similrly n prove tht (try it!) n thus G( + h) G(x) lim = lim h 0 h h 0 h G( + h) G(x) lim = lim h 0 h h 0 h +h +h +h f(x) = f(). f(x) = f(). f(x) = f(). We n lso similrly prove the sme if f() is negtive. Prt II of the FTC: Assume tht F (x) ny ntierivtive of f(x) n ssume tht G(x) is the funtion G(x) = x f(t)t, for x b. Then, by prt I of the FTC, G(x) is lso n ntierivtive of f(x). Thus by Theorem of this setion, it follows tht G(x) = F (x)+c for some onstnt C. Therefore (note tht G() = 0) b f(t)t = G(b) = G(b) G() = (F (b) + C) (F () + C) = F (b) F ().
Given the lose reltionship between efinite integrl (whih is limit of Riemnn sums) n ntierivtive we enote the set of ntierivtives of ontinuous funion y = f(x) by f(x). 5 So, for exmple sin(x) = os(x) + C. Exmple. Using the FTC () (b) () () sin(x 2 ) = sin(x2 ) = π/2 0 sin(x2 ) = π/2 sin(x 2 ) = 0
6 Tble of bsi ntierivtives: f(x) x r, x e x sin(x) os(x) se 2 (x) s 2 (x) r se(x) tn(x) s(x) ot(x) x 2 x 2 f(x) r + xr+ + C ln x + C e x + C os(x) + C sin(x) + C tn(x) + C ot(x) + C se(x) + C s(x) + C rsin(x) ros(x) + x 2 rtn(x)
7 Problems: Problem. Fin ll ntierivtives of: () 5 sin(x) + x 5 7, (b) e x + x 2.
8 Problem 2. Prove tht F (x) = x ln(x) x + C, x > 0, is n ntierivtive of ln(x).
9 Problem 3. Evlute the following integrls: ( ) 3 ) x 3 3 x b) ) 2 0 π 2 ( 3e x 2 x), f(x), where x 2 if 2 x < 0, f(x) = sin(x) if x π 2.
0 Problem 4. Compute: ) b) ) x 2 2 5 ln 2 (x) e t2 t, rtn(t)t, e x 2 e t2 sin(t)t. e x
Problem 5. Evlute: 3 0 x 2 4.