Fundmentl Theorem of Clculus. Prt I Connection between integrtion nd differentition. Tody we will discuss reltionship between two mjor concepts of Clculus: integrtion nd differentition. We will show tht these opertions re inverse to ech other. We will do so by defining structure tht llows to recover the function from its derivtive using definite integrl. Let me proceed to the first slide. Slide Motivtion: The problem of finding ntiderivtives. Consider the following question: Given function, we cn find its derivtive, or differentite it. Now, given the derivtive, cn we find the function bck? Cn we ntidifferenitte it? 2 Slide 2 Definition. An ntiderivtive of function f() is function F () such tht F () = f(). In other words, given the function f(), you wnt to tell whose derivtive it is. Emple. Find n ntiderivtive of. The nswer: An ntiderivtive of is. Check by differentition. Emple 2. Find n ntiderivtive of + 2.
The nswer: An ntiderivtive of is rctn. + 2 3 Slide 3 How do you know? 4 Slide 4 Some ntiderivtives cn be found by reding differentition formuls bckwrds. Indeed, ccording to the formuls from clculus book, is the derivtive of rctn + 2 5 Slide 5 However, no clculus book hs formuls for sin( 2 ), e (2), + sin 2 ( 3 ) The question rises: do these functions hve ntiderivtives? 6 Slide 6 An observtion: No mtter wht object s velocity v(t) is, its position function d(t) is lwys n ntiderivtive of v(t), tht is d (t) = v(t). This suggests tht ll functions hve ntiderivtives. 2
7 Slide 7 Suppose the speed of my cr obeys sin(t 2 ) (do not try it on the rod!). The cr will move ccordingly nd the position of the cr F (t) will give the ntiderivtive of sin(t 2 ). 8 Slide 8 A hypothesis: Clculting ntiderivtives must be similr to clculting position from velocity. 9 Slide 9 We rrived to the second prt of our discussion: nive derivtion of Fundmentl Theorem of Clculus. 0 Slide 0 Let, t initil time t 0, position of the cr on the rod is d(t 0 ) nd velocity is v(t 0 ). LET ME SWITCH TO OVERHEAD TO MAKE A QUICK COMMENT: At moment t 0, velocity of the cr is v(t 0 ). During the period of time t the cr will trvel pproimtely v(t 0 ) t. Thus, the new position of the cr is d(t ) d(t 0 ) + v(t 0 ) t, (t = t 0 + t) 3
Slide Similrly, t time t the velocity of the cr is v(t ). During the period t the cr will trvel pproimtely v(t ) t. Position of the cr fter two moments of time is d(t 2 ) d(t 0 ) + v(t 0 ) t + v(t ) t, (t 2 = t + t) 2 Slide 2 Similrly, position of the cr fter n moments of time is d(t n ) d(t 0 ) + v(t 0 ) t +... + v(t n ) t which cn be written shorter using sigm nottion: = d(t 0 ) + n i=0 v(t i ) t LET ME GO BACK TO COMPUTER SCREEN: 3 Slide 3 Now let the number n of time steps before we rech certin moment of time t increses infinitely, which mens tht the size of time step t decreses to 0; the epression becomes limit: n d(t) d(t 0 ) = lim v(t i ) t t 0 i=0 Compre this limit to the definition on the definite integrl: t t 0 v(τ) dτ = lim t 0 4 n i=0 v(t i ) t
Epressions in red coincide, therefore 4 Slide 4 d(t) d(t 0 ) = v(τ) dτ t 0 while d(t) is being position function corresponding to velocity v(t), tht is d (t) = v(t). t 5 Slide 5 Assuming t 0 = 0, d(t 0 ) = 0, distnce cn be clculted from velocity by d(t) = t 0 v(τ) dτ First of ll, is this function in our regulr sense? Well, Yes. For ech vlue t it defines unique number d(t) Then if d (t) = v(t)? Agin, Yes. This constitutes the ssertion of Fundmentl Theorem of Clculus. 6 Slide 6 Fundmentl Theorem of Clculus. Prt I. Let f() be continuous function (so, the definite integrl of f() eists). Then the function F () = f(τ) dτ. is n ntiderivtive of f(), which is tht F () = f(). 5
7 Slide 7 EXAMPLES 8 Slide 8 Consider n emple of evluting n ntiderivtive of function f() = sin( 2 ). I WILL SWITCH TO OVERHEAD NOW. According to the Fundmentl Theorem of Clculus, Prt I, the function F () = is n ntiderivtive of f() = sin( 2 ). 0 sin(τ 2 ) dτ Indeed, ccording to the formul in the upper right corner, derivtive of F () is sin( 2 ) 9 Slide 9 Emple 4. Find n ntiderivtive of f() = e (2). Agin, we use the formul in upper right corner: substituting e (τ 2) for f(τ) we obtin n ntiderivtive of e (2 ) G() = e (τ 2) dτ 0 6
20 Slide 20 2 Slide 2 22 Slide 22 23 Slide 23 24 Slide 24 25 Slide 25 26 Slide 26 27 Slide 27 28 Slide 28 29 Slide 29 30 Slide 30 3 Slide 3 32 Slide 32 7 We rrived to the lst prt of our discussion: the proof of Fundmentl Theorem of Clculus. Prt I.
LET ME SWITCH TO THE COMPUTER SCREEN 33 Slide 33 We will be using two fcts: Intervl Additive Property: b f(τ) dτ = c f(τ) dτ + nd Comprison Property: If m f() M on [, b] m(b ) b b c f(τ) dτ f(τ) dτ M(b ) 34 Slide 34 Let F () = f(τ) dτ. We will now show tht F () = f(). By definition of derivtive, F F ( + h) F () () = lim h 0 h = lim h 0 f(τ) dτ f(τ) dτ h 35 Slide 35 Notice, tht by the Intervl Additive Property, the epression in the numertor cn be simplified: f(τ) dτ f(τ) dτ = 8 f(τ) dτ
Therefore, F () = lim h 0 = lim h 0 h f(τ) dτ f(τ) dτ h f(τ) dτ 36 Slide 36 Finlly, if we define two numbers: m = Then the obvious inequlity holds min f(τ), M = m f(τ) τ [,+h] τ [,+h] or, dividing by h, mh m h f(τ) dτ Mh f(τ) dτ M 37 Slide 37 Now by s we shrink the intervl [, + h] by considering limit s h 0. Both m nd M converge (due to the continuity of f()) to the vlue of f(). Therefore, by the Squeeze Theorem, the epression converges to which implies tht The theorem is proved. m h f(τ) dτ M +h f() lim f(τ) dτ f() h 0 h +h lim f(τ) dτ = f() h 0 h 9