Fundamental Theorem of Calculus

Transcription

1 Fundamental Theorem of Calculus MATH 6 Calculus I J. Robert Buchanan Department of Mathematics Summer 208

2 Remarks The Fundamental Theorem of Calculus (FTC) will make the evaluation of definite integrals more convenient. The FTC unifies the topics of definite integrals and derivatives.

3 Fundamental Theorem of Calculus, Part I Theorem (FTC, Part I) If f is continuous on [a, b] and F() is any antiderivative of f (), then b a f () d = [F()] =b =a = F(b) F(a).

4 Fundamental Theorem of Calculus, Part I Theorem (FTC, Part I) If f is continuous on [a, b] and F() is any antiderivative of f (), then b Alternative notations: b a a f () d = [F()] =b =a = F(b) F(a). f () d = F(b) F(a) = [F()] =b =a = F() b a = [F()]b a = F()]b a.

5 Proof of FTC, Part I ( of 2) Let a = 0 < < < n = b be a regular partition of [a, b]. F(b) F(a) = F( n ) F( 0 ) = F( n ) F( n ) + F( n ) F ( n 2 ) + + F( 2 ) F( ) + F( ) F( 0 )

6 Proof of FTC, Part I ( of 2) Let a = 0 < < < n = b be a regular partition of [a, b]. F(b) F(a) = F( n ) F( 0 ) = F( n ) F( n ) + F( n ) F ( n 2 ) + + F( 2 ) F( ) + F( ) F( 0 ) = (F( n ) F( n )) + (F( n ) F( n 2 )) + + (F( 2 ) F( )) + (F( ) F( 0 ))

7 Proof of FTC, Part I ( of 2) Let a = 0 < < < n = b be a regular partition of [a, b]. F(b) F(a) = F( n ) F( 0 ) = F( n ) F( n ) + F( n ) F ( n 2 ) + + F( 2 ) F( ) + F( ) F( 0 ) = (F( n ) F( n )) + (F( n ) F( n 2 )) + + (F( 2 ) F( )) + (F( ) F( 0 )) n = (F( i ) F( i )) = = i= n F (c i )( i i ) i= n f (c i )( i i ) i= (by the MVT)

8 Proof of FTC, Part I (2 of 2) F(b) F(a) = n f (c i ) i= = lim n = b a n f (c i ) i= f () d

9 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals ( 2 2) d ( 3 2/3 ) d d

10 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 2 [ ] =2 ( 2 2) d = = ( 3 2/3 ) d d =

11 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 2 [ ] =2 ( 2 2) d = = 3 = 27 ( 3 [ 3 2/3 ) d = 4 4/3 3 ] =27 5 5/3 = d =0

12 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 2 [ ] =2 ( 2 2) d = = 3 = 27 ( 3 [ 3 2/3 ) d = 4 4/3 3 ] =27 5 5/3 = [ + 2 d = 4 tan ] = = = 2π =0

13 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 π/ (e e ) d 2 cos 2 d d

14 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 [ d = 3 ln 4 ] =3 = 5 ( ) 3 =2 3 3 ln 2 π/4 0 (e e ) d 2 cos 2 d

15 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 [ d = 3 ln 4 ] =3 = 5 ( ) 3 =2 3 3 ln 2 π/4 0 (e e ) d = [ e + e ] = = = 0 2 cos 2 d

16 Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 [ d = 3 ln 4 ] =3 = 5 ( ) 3 =2 3 3 ln 2 π/4 0 (e e ) d = [ e + e ] = = = 0 2 cos 2 d = [2 tan ]=π/4 =0 = 2

17 Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ (t 2 + 2t) ] t= t= =

18 Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ ] t= (t 2 + 2t) = t= Note: If F() = then F () = f (), in other words [ d ] (2t + 2) dt = f () d or (2t + 2) dt is an antiderivative of f ().

19 Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ ] t= (t 2 + 2t) = t= Note: If F() = then F () = f (), in other words [ d ] (2t + 2) dt = f () d or The function (2t + 2) dt is an antiderivative of f (). (2t + 2) dt gives the net area under the graph of f (t) on the interval [, ].

20 Fundamental Theorem of Calculus, Part II Theorem (FTC, Part II) If f is continuous on [a, b] and F() = F () = f () on [a, b]. a f (t) dt, then

21 Proof of FTC, Part II F F( + h) F() () = lim h 0 h [ +h = lim h 0 h a [ = lim h 0 h a +h h 0 h = lim = lim f (c) h 0 = f () f (t) dt f (t) dt a f (t) dt + f (t) dt +h ] ] f (t) dt f (t) dt a (with c + h by IMVT)

22 Eamples Find the derivatives of the following functions using the FTC, Part II and other derivative formulas as appropriate. F() = 2 (t 2 4t 3) dt G() = (t 2 4t 3) dt H() = 2 sin t dt

23 Eamples Find the derivatives of the following functions using the FTC, Part II and other derivative formulas as appropriate. F() = 2 (t 2 4t 3) dt F () = G() = (t 2 4t 3) dt H() = 2 sin t dt

24 Eamples Find the derivatives of the following functions using the FTC, Part II and other derivative formulas as appropriate. F() = 2 (t 2 4t 3) dt F () = G() = H() = (t 2 4t 3) dt G () = (( 2 + ) 2 4( 2 + ) 3)( 2 + ) = 2(( 2 + ) 2 4( 2 + ) 3) 2 sin t dt

25 Eamples Find the derivatives of the following functions using the FTC, Part II and other derivative formulas as appropriate. F() = 2 (t 2 4t 3) dt F () = G() = (t 2 4t 3) dt G () = (( 2 + ) 2 4( 2 + ) 3)( 2 + ) = 2(( 2 + ) 2 4( 2 + ) 3) H() = 2 sin t dt H () = d d [ ] sin t dt = sin 2

26 Eample Find the derivative of the following function using the FTC, Part II and other derivative formulas as appropriate. J() = 2 +2 (t 2 4t 3) dt

27 Eample Find the derivative of the following function using the FTC, Part II and other derivative formulas as appropriate. J() = = 2 +2 = (t 2 4t 3) dt (t 2 4t 3) dt + (t 2 4t 3) dt (t 2 4t 3) dt (t 2 4t 3) dt

28 Eample Find the derivative of the following function using the FTC, Part II and other derivative formulas as appropriate. J() = = 2 +2 = (t 2 4t 3) dt (t 2 4t 3) dt + (t 2 4t 3) dt (t 2 4t 3) dt (t 2 4t 3) dt J () = ( 2 4 3) + (( 2 + 2) 2 4( 2 + 2) 3)( 2 + 2) = (( 2 + 2) 2 4( 2 + 2) 3)

29 Eample Find the equation of the tangent line to the graph of f () = at the point where =. ln(t 2 + 2t + ) dt

30 Solution Point of tangency: ( (, f ()) =, ln(t 2 + 2t + ) dt ) = (, 0)

31 Solution Point of tangency: ( (, f ()) =, Slope of tangent line: ln(t 2 + 2t + ) dt f () = ln( ) f () = ln 4 ) = (, 0)

32 Solution Point of tangency: ( (, f ()) =, Slope of tangent line: ln(t 2 + 2t + ) dt f () = ln( ) f () = ln 4 Equation of the tangent line: m = y y 0 0 ln 4 = y 0 y = (ln 4)( ) ) = (, 0)

33 Graph f () = ln(t 2 + 2t + ) dt y = (ln 4)( ) y

34 Homework Read Section 4.5 Eercises: 6 odd