MATH 12002 - CALCULUS I 4.4: Fundamental Theorem of Calculus - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 4.4: Fundamental Theorem of Calculus Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 8

The Fundamental Theorem Fundamental Theorem of Calculus Let f be continuous on [ a , b ] . 1 If we define � x g ( x ) = f ( t ) dt , a then g ′ ( x ) = f ( x ) for all x in ( a , b ) . 2 If F is any antiderivative for f , then � b f ( x ) dx = F ( b ) − F ( a ) . a D.L. White (Kent State University) 2 / 8

The Fundamental Theorem Notes: Part 1 says that the function g is an antiderivative of f . In particular, every continuous function has an antiderivative. � x Part 1 also says that if we integrate f (i.e., to get a f ( t ) dt ) and then differentiate, we get f back; that is, � x d f ( t ) dt = f ( x ) . dx a Part 2 says that if we differentiate F (i.e., to get f ) and then integrate, we get F back (except for a constant); that is, � x F ′ ( t ) dt = F ( x ) − F ( a ) . a Together, Parts 1 & 2 say that differentiation and integration are inverse processes. That Fundamental Theorem really ties the branches of calculus together, does it not? D.L. White (Kent State University) 3 / 8

Proof of FTC � x Assume f is continuous on [ a , b ] and define g ( x ) = a f ( t ) dt . For Part(1), we need to show that g ′ ( x ) = f ( x ) for a < x < b ; that is, g ( x + h ) − g ( x ) lim = f ( x ). h h → 0 We first informally outline the idea of the proof of Part (1). � x + h � x Note that g ( x + h ) = f ( t ) dt and g ( x ) = a f ( t ) dt . a For a positive function f and positive h , we have the following picture: ✻ R x + h R x f ( x + h ) g ( x + h ) − g ( x ) = f ( t ) dt − a f ( t ) dt f q a = area under graph on [ x , x + h ] f ( x ) q ≈ area of rectangle f ( x ) ✲ h = h · f ( x ) a x x + h b Hence for small h , g ( x + h ) − g ( x ) g ( x + h ) − g ( x ) ≈ f ( x ) and so f ( x ) = lim = g ′ ( x ). h h h → 0 D.L. White (Kent State University) 4 / 8

Proof of FTC We now prove Part (1) more rigorously. By one of our properties of definite integrals, � x + h g ( x + h ) = f ( t ) dt a � x � x + h = a f ( t ) dt + f ( t ) dt x � x + h = g ( x ) + f ( t ) dt , x and so � x + h g ( x + h ) − g ( x ) = f ( t ) dt . x Therefore, � x + h g ( x + h ) − g ( x ) = 1 f ( t ) dt . h h x D.L. White (Kent State University) 5 / 8

Proof of FTC Since f is continuous on the closed interval [ x , x + h ], it has an absolute maximum and an absolute minimum on the interval. Thus there is a number u , x � u � x + h , such that f ( u ) is the absolute minimum value of f on the interval, and there is a number v , x � v � x + h , such that f ( v ) is the absolute maximum value of f on the interval. Hence on the interval [ x , x + h ], we have f ( u ) � f ( t ) � f ( v ). By a comparison property of integrals, we have � x + h f ( u )(( x + h ) − x ) � f ( t ) dt � f ( v )(( x + h ) − x ); x that is, � x + h f ( u ) h � f ( t ) dt � f ( v ) h , x and finally � x + h f ( u ) � 1 f ( t ) dt � f ( v ). h x D.L. White (Kent State University) 6 / 8

Proof of FTC Recalling that � x + h � x + h f ( t ) dt = g ( x + h ) − g ( x ) 1 and f ( u ) � 1 f ( t ) dt � f ( v ), h x h h x we now have f ( u ) � g ( x + h ) − g ( x ) � f ( v ). h Since x � u � x + h & x � v � x + h , we have u → x & v → x as h → 0. Since f is continuous, this implies h → 0 f ( u ) = lim lim u → x f ( u ) = f ( x ) and lim h → 0 f ( v ) = lim v → x f ( v ) = f ( x ). Therefore, g ( x + h ) − g ( x ) f ( x ) = lim h → 0 f ( u ) � lim � lim h → 0 f ( v ) = f ( x ), h h → 0 and so by the Squeeze Theorem g ( x + h ) − g ( x ) g ′ ( x ) = lim = f ( x ). h h → 0 D.L. White (Kent State University) 7 / 8

Proof of FTC We now have that Part (1) of the Fundamental Theorem holds, � x and so g ( x ) = a f ( t ) dt is an antiderivative of f . If F is any antiderivative of f , then F ( x ) = g ( x ) + C for some constant C . In particular, � b F ( b ) = g ( b ) + C = f ( t ) dt + C , a � b and so a f ( t ) dt = F ( b ) − C . But also, � a F ( a ) = g ( a ) + C = f ( t ) dt + C = 0 + C = C , a and so C = F ( a ). Therefore � b f ( t ) dt = F ( b ) − F ( a ) a and Part (2) of the Fundamental Theorem holds. D.L. White (Kent State University) 8 / 8