The Fundamental Theorem of Calculus


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1 The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018
2 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls using the Fundmentl Theorem of Clculus. Find the verge vlue of function over n intervl.
3 Grphicl Ide Consider the shded re R beneth the grph of y = f (x), bove the xxis, nd between x = nd x = b. y y f x R b
4 Definite Integrl Definition Let f (x) 0 nd continuous on the closed intervl [, b]. The re of the region bounded by the grph of f, the xxis, nd the lines x = nd x = b is denoted by The expression Are = f (x) dx. f (x) dx is clled the definite integrl from to b, where is clled the lower limit of integrtion nd b is clled the upper limit of integrtion.
5 Exmple (1 of 2) Evlute 2 0 3x dx using the grph below. y
6 Exmple (2 of 2) Evlute 2 1 x 1 dx using the grph below. 3.0 y
7 Generl Regions Beneth Curves For regions tht re not fmilir geometric figures, we need to pproximte the re using rectngulr strips nd summtion of their res. y b
8 Riemnn Sum (1 of 3) Prtition the intervl [, b] into n subintervls of length x = b n. x x x x x The endpoints of the intervls re the vlues x k = + k x for k = 0, 1,..., n.
9 Riemnn Sum (2 of 3) Choose n evlution point c k in ech subintervl [x k 1, x k ]. c 1, f c 1 c 2, f c 2 c 3, f c 3 c 4, f c 4 x x x x x The midpoints ( of the intervls re the vlues c k = + k 1 ) x for k = 1,..., n. 2
10 Riemnn Sum (3 of 3) Approximte the re under the curve in subintervl [x k 1, x k ] by the re of the rectngle whose bse is x nd whose height is f (c k ). c 1, f c 1 c 2, f c 2 c 3, f c 3 c 4, f c 4 x x x x x
11 Generl Form of Riemnn Sum Definition The generl form of Riemnn sum for function y = f (x) which is continuous on [, b] is S n = [f (c 1 ) + f (c 2 ) + + f (c n )] x where x = b, n is the number of subintervls of [, b], n nd c 1, c 2,..., c n represents one xvlue from ech subintervl.
12 Exmple Find the Riemnn sum S 5 for f (x) = 9 x 2 on [ 3, 2] with c 1 = 2.5, c 2 = 1.5, c 3 = 0.5, c 4 = 0.5, nd c 5 = 1.5.
13 Exmple Find the Riemnn sum S 5 for f (x) = 9 x 2 on [ 3, 2] with c 1 = 2.5, c 2 = 1.5, c 3 = 0.5, c 4 = 0.5, nd c 5 = 1.5. x = 2 ( 3) 5 = 5 5 = 1
14 Exmple Find the Riemnn sum S 5 for f (x) = 9 x 2 on [ 3, 2] with c 1 = 2.5, c 2 = 1.5, c 3 = 0.5, c 4 = 0.5, nd c 5 = 1.5. x = 2 ( 3) = = 1 S 5 = [f (c 1 ) + f (c 2 ) + f (c 3 ) + f (c 4 ) + f (c 5 )] x = [f ( 2.5) + f ( 1.5) + f ( 0.5) + f (0.5) + f (1.5)] (1) = S 5 = 33.75
15 Are Under Curve Definition If function y = f (x) is nonnegtive nd continuous on intervl [, b], then the re under the curve is defined to be A = lim n S n, where S n is the generl form of Riemnn sum for the function f.
16 Exmple Estimte the re under the curve f (x) = 9 x 2 on the intervl [ 3, 2] using the limit of the Riemnn sum.
17 Exmple Estimte the re under the curve f (x) = 9 x 2 on the intervl [ 3, 2] using the limit of the Riemnn sum. n S n
18 Comments 1. We hve ssumed tht f (x) 0 on [, b], but this is not necessry. However, if f (x) < 0 for some x in [, b] the Riemnn sum pproximtes the net re bove the xxis. 2. We hve ssumed tht ll of the subintervls re of the sme length. Agin, this is not necessry. We my mix subintervls of different lengths in our clcultion of Riemnn sum.
19 The Definite Integrl Definition If function y = f (x) is continuous on the intervl [, b], then the definite integrl of f from to b is denoted is defined to be f (x) dx = lim n S n, f (x) dx nd where S n is the generl form of Riemnn sum for the function f. The number is clled the lower limit of integrtion nd the number b is clled the upper limit of integrtion.
20 The Definite Integrl Definition If function y = f (x) is continuous on the intervl [, b], then the definite integrl of f from to b is denoted is defined to be f (x) dx = lim n S n, f (x) dx nd where S n is the generl form of Riemnn sum for the function f. The number is clled the lower limit of integrtion nd the number b is clled the upper limit of integrtion. If f (x) 0 on intervl [, b] then the re under the curve is A = f (x) dx.
21 Fundmentl Theorem of Clculus The most convenient method for evluting definite integrls is to use the result below. Theorem (Fundmentl Theorem of Clculus) If y = f (x) nd continuous on the closed intervl [, b], then f (x) dx = F(b) F () where F(x) is ny ntiderivtive of f (x) for ll x in [, b].
22 Fundmentl Theorem of Clculus The most convenient method for evluting definite integrls is to use the result below. Theorem (Fundmentl Theorem of Clculus) If y = f (x) nd continuous on the closed intervl [, b], then f (x) dx = F(b) F () where F(x) is ny ntiderivtive of f (x) for ll x in [, b]. Alterntive nottion: f (x) dx = [F(x)] b = F(b) F()
23 Properties of Definite Integrls f (x) dx = 0 k f (x) dx = k [f (x) ± g(x)] dx = f (x) dx f (x) dx ± g(x) dx
24 Exmple (1 of 2) Use the Fundmentl Theorem of Clculus to evlute the definite integrl 3 2 (x 2 + 1) dx.
25 Exmple (1 of 2) Use the Fundmentl Theorem of Clculus to evlute the definite integrl 3 2 (x 2 + 1) dx. 3 2 (x 2 + 1) dx = = = 22 3 [ ] x 3 + x 2 ( ) 1 3 (3)3 + 3 ( ) 1 3 (2)3 + 2
26 Exmple (2 of 2) Use the Fundmentl Theorem of Clculus to evlute the definite integrl 1 0 (2x + 3) 3 dx.
27 Exmple (2 of 2) Use the Fundmentl Theorem of Clculus to evlute the definite integrl 1 0 (2x + 3) 3 dx (2x + 3) 3 dx = 1 (2x + 3) 3 (2) dx 2 0 [ ] 1 1 = (2x + 3)4 8 0 = 1 [(2(1) + 3) 4 (2(0) + 3) 4] 8 = 1 [625 81] 8 = = 68
28 Averge Vlue of Function Definition If f (x) is continuous on [, b], then the verge vlue of f on [, b] is AV = 1 f (x) dx b
29 Exmple The cost per unit c for producing rollerbldes over the next 24 months is modeled by the function c = 0.005t t where 0 t 24. Find the verge cost per unit over the intervl [0, 24]. c
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