3.4 The Fundamental Theorem of Calculus (FTC) The Mean Value Theorem of Integrals And Average Value * The Mean Value Theorem (p214): If Fis continuous on [],a band diff’able on the (),a b Then there exists a (),c a b∈such that ( )()()'F b F aF cb a−=− a c b * The Fundamental Theorem of Calculus (part 1): If fis continuous on [],a band Fis any antiderivative of fon [],a b Then ( ) ( ) ( )baf x dx F b F a= −∫ f a b x0 xn Proof: Let 1 2 1...na x x x b−< < < < <be any partition of [],a b (This partition divides the [],a binto nsubintervals: [][][][][]1 1 2 2 3 2 1 1, ; , ; , ;... , ; ,n n na x x x x x x x x b− − −) Whose widths are… ∆x1 ∆x2 ∆x3 ∆xn-1 ∆xn :By hypothesis ()()'F x f x=for all x in [],a b So Fsatisfies the hypothesis of the MVT on each subinterval Thus we can find numbers * * * *1 2 3, , ,...,nx x x xin the respective subintervals such that ( )()()( )( ) ( )( )( ) ( )1*112 1*22 11*1'''n nnn nF x F aF xx aF x F xF xx xF x F xF xx x−−−=−−=−−=−“OR” ( )1F x()F a− = ()()*1 1'F x x a⋅ − = ()*1 1f x x⋅ ∆ ( )2F x()1F x−= ()()*2 2 1'F x x x⋅ − = ()*2 2f x x⋅ ∆ ( )3F x()2F x−= ()()*3 3 2'F x x x⋅ − = ()*3 3f x x⋅ ∆ ADD  ( )1nF x−( )2nF x−−()()*1 1 2'n n nF x x x− − −= ⋅ − = ()*1 1n nf x x− −⋅ ∆ ( ) ( )1nF b F x−−= ()()*1'n nF x b x−⋅ − = ()*n nf x x⋅ ∆ ()()F b F a− = ( )*1nk kkf x x=∆∑ * Now increase n in such a way that the largest ∆xk gets smaller. ( ) ( )( )*1nk kkF b F a f x x=− = ∆∑ ( ) ( )( )( )*10 0lim limnk kkF b F a f x x∆ → ∆=→− = ∆∑ ( ) ( )( )( )*1lim limnk kn nkF b F a f x x→∞ →∞=− = ∆∑ By definition of definite intergral… ( ) ( ) ( )baF b F a f x dx− =∫ Examples: ]112 3001713x dx x= +∫ ( ) ( )3 31 11 71 0 713 3   = + − +       1 171 0 713 3= + − − = * Note that we chose the constant to be 71, but we could have chosen C to be any number as it would have cancelled just like you see the 71’s cancelling in general we will choose c=0. 3300sin cos 0xdx xππ= − +∫ ( )cos cos03π= − − − 1 112 2= − + =Fundamental Theorem of Calculus (FTC) II * If fis continuous on an intervalI, then fhas an antiderivative onI In particular, if ais any number in I Then the function Fdefined by ( ) ( )xaF x f t dt=∫ Is an antiderivative of Fon I; that is ()()'F x f x=for allxin I Another way to write is: ( ) ( )xadf t dt f xdx =  ∫ Example: 1) Find 5cos cosxdtdt xdx =  ∫ [ ]55cos sinxxd dtdt tdx dx =  ∫ [ ]sin sin 5dxdx= − cos 0 cosx x= − = 2) ( )223cos 2 cosxdtdt x xdx =   ∫ 23sinxdtdx    2sin sin 3dxdx = −  2 22 cos 0 2 cosx x x x= + = *Note: If upper limit is more than just X then need a “chain rule” type piece.3) ( )11 2xF x t dt= − −∫ For what values of xis ()0F x= ( ) ( ) ( )1 5 31 1 11 0; 5 0; 3 0F F F−= = = = − = =∫ ∫ ∫ 1 2y t= − − * Note that this type of function is a net sigh area accumulator. ( )2132 1 22F t dt= − − = −∫ ( )0130 1 22F t dt= − − =∫ ************************************************************************************************************ MVT For Integrals: If fis continuous on [],a bthen there exists an *xin [],a bsuch that ( )( )( )*baf x dx f x b a= ⋅ −∫ max min a x* b ()()Ab aa Mm b≤ ≤−−Example: 12013x dx=∫ ()*f xis called the “average value” of fon [],a b ( )( )*1baf x f x dxa b=−∫ 1 Work: ( )1* 201 1 11 0 1 3f x x dx= = ⋅−∫ ( )*13f x= 32 1 ( )2* *1 1 1 3.5773 3 33x x or= = = =