Fundmentl Theorem of Clculus
Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under the grph of f from to. Then the following fcts re cler or were mentioned previously.. A () = f(). A() = 0 3. A() = A
We illustrte the proof of. A(+h)
A()
A( + h) - A(h) A ( + h) A () h f() if h is smll, so f() = lim A ( + h) A () = A () h h
Thus A is n ntiderivtive of f. Now suppose tht F is ny other ntiderivtive of f. Then, for some constnt C, we hve F () = A () + C Then F () F () = A () + C A () + C = A () A () = f() d This result continues to hold for functions tht re integrle, whether or not they re positive, Thus we hve Theorem. (The fundmentl Th. of Clculus, Prt ). If f is continuous on [, ] nd if F is ny ntiderivtive of f, then f () d = F () F ()
Proof. Choose points,,, n in [, ] so tht < < < < n < These points divide [, ] into n equl length suintervls [, ], [, ],, [ n-, ] whose lengths re. We hve ssumed tht F () = f() in ech intervl, so y the men vlue theorem, we cn find points *, *,, n * in the respective intervls so tht F ( ) F () = f( * ) F ( ) F ( ) = f( * ) F () F ( n ) = f( * ) n n
F ( ) F () = f( * ) F ( ) F ( ) = f( * ) F () F ( ) = ( *) n f n Adding these equtions produces n F () F () = f( * ) k k= The left hnd side is independent of n nd the right hnd side will tend to f () d tend to 0. k s n goes to infinity nd the
We use the nottion F () = F () F (). Thus we cn write the fundmentl theorem s: If F () = f(), then f() d= F () = F () F (). Emple. Use the fundmentl theorem of Clculus to find 3 d 3 Solution. The function F ()= hs s derivtive. So 3 3 3 3 7 6 d= = = 3 3 3 3
The Reltionship etween the Definite nd Indefinite Integrl Let f e continuous in [, ]. We hve seen tht if F nd G re ny two ntiderivtives, tht is ny two functions in the set f() d then: F () G () f() d = =
We denote this fct y writing f () d = f () d Thus we summrize s follows: To find the definite integrl of continuous function f from to, we choose ny function F in the indefinite integrl (ny ntiderivtive), nd evlute F() F(). Of course, this will only work when we know the formul for n ntiderivtive.
Emple: Using the fundmentl theorem, find cos( d ) Solution. We know from sic formuls tht cos( d ) = sin( ) + C. π π Thus π cos( d ) = sin( ) = sin sin(0) = 0= 0 0 Emple: Using the fundmentl theorem, find 5 4 3 d Solution. 5 5 5 5 5 4 3 d d 4d 3 4 = = = 5 4 65 6 = 609= 588 ( ) ( ) π 0
Emple: Find the re under the curve etween nd 4. f() = 3 5 3 5 5 5 Solution. We know from sic formuls tht d= + C Thus 4 3 4 5 5 5 5 5 5 d 4 5 5 6 = = =
Emple: Find the under the curve f() = etween nd 5. Solution. We know from sic formuls tht Thus 5 5 d = d ln ln(5) = = d= ln( ) + C
Emple: Evlute the definite integrl ln7 4 e d ln4 Solution. We know from sic formuls tht e d= e+ C Thus ln(7) ln(7) ln(7) 4e d 4 e d 4e 4 e ln(7) e ln(4) = = = = 43 = ln(4) ln(4) ln(4)
Emple: Evlute the definite integrl ( ) + e csccot d Solution. ( ) + e csccot d = d+ e d+ ( csccot d ) = ln + e + csc( ) ln() + ( e e) + csc() csc()
Emple: Evlute the definite integrl 3π 4 cos( d ) 0 y = cos() Solution. 3π π 3π 4 4 cos( d ) = cos( d ) + cos( d ) 0 0 π π 3π 4 π π 3π 4 = cos( ) cos( ) sin( ) sin( ) ( 0) d d= = = π 0 0
Theorem. (Men Vlue Theorem for Integrls) If f is continuous on closed intervl [, ] then there is t lest one numer * in [, ] for which f () d = f (*)( ) Proof. Since f is continuous, we know tht it chieves mimum vlue M nd minimum vlue m in [, ]. Thus for ll in [, ] we hve m f() M, nd therefore m ( ) = md f() d Md= M ( ) or m f() d M ( )
Thus the numer y = f() d 0 lies etween the ( ) mimum nd minimum vlue of f on [, ]. By the intermedite vlue theorem, there must e point * so tht (*) f = y = f() d 0 ( ) f () d ( ) y = 0 *
We use this result to prove the following theorem. Theorem. (The Fundmentl Theorem of Clculus, Prt ). If f is continuous function on n intervl I, then f hs n ntiderivtive on I. In prticulr if we define F y F () = ftdt () then F () = f(). Proof. + h + h F ( + h) F () = f() tdt ftdt () = ftdt () = hf(*) t for some point t h * etween nd + h. Then F () = lim F ( + h) F () = lim ft (*) = f() h 0 h h 0 h
Emple. () d 4 t dt = 4 d 3 () d cos( t) cos( ) dt = d 3 t Ech of these fcts follows from the second prt of the fundmentl theorem of clculus. Sometimes such fct cn lso e proved independently. For emple we see tht 5 4 t 5 43 t dt = =. Then we cn compute the derivtive 3 5 5 5 3 directly nd it is 4. However, we cnnot tret the second emple in this wy since we do not know the formul for n ntiderivtive.