Fundamental Theorem of Calculus

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

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.

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).

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.

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 )

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 ))

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

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

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

Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 2 [ ] =2 ( 2 2) d = 3 3 2 = 3 = 27 ( 3 [ 3 2/3 ) d = 4 4/3 3 ] =27 5 5/3 = 70 20 0 4 + 2 d =0

Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 2 [ ] =2 ( 2 2) d = 3 3 2 = 3 = 27 ( 3 [ 3 2/3 ) d = 4 4/3 3 ] =27 5 5/3 = 70 20 0 4 [ + 2 d = 4 tan ] = = = 2π =0

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

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

Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 [ 3 + 4 2 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

Eamples Use the Fundamental Theorem of Calculus, Part I to evaluate the following definite integrals. 3 2 [ 3 + 4 2 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

Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ (t 2 + 2t) ] t= t= = 2 + 2 3

Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ ] t= (t 2 + 2t) = 2 + 2 3 t= Note: If F() = 2 + 2 3 then F () = f (), in other words [ d ] (2t + 2) dt = f () d or (2t + 2) dt is an antiderivative of f ().

Net Area Function Suppose f (t) = 2t + 2 on the interval [, ). (2t + 2) dt = [ ] t= (t 2 + 2t) = 2 + 2 3 t= Note: If F() = 2 + 2 3 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 [, ].

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

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)

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() = 2 + 2 (t 2 4t 3) dt H() = 2 sin t dt

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 () = 2 4 3 G() = 2 + 2 (t 2 4t 3) dt H() = 2 sin t dt

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 () = 2 4 3 G() = H() = 2 + 2 (t 2 4t 3) dt G () = (( 2 + ) 2 4( 2 + ) 3)( 2 + ) = 2(( 2 + ) 2 4( 2 + ) 3) 2 sin t dt

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 () = 2 4 3 G() = 2 + 2 (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

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

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 + 2 +2 2 +2 (t 2 4t 3) dt (t 2 4t 3) dt

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

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

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

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

Graph f () = ln(t 2 + 2t + ) dt y = (ln 4)( ) y 2.0.5.0 0.5 0.5.0.5 2.0-0.5 -.0 -.5

Homework Read Section 4.5 Eercises: 6 odd