5: The Definite Integral

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1 5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce tht the oject hs ( ) trveled during the first four seconds? The old d = r t only wors for constnt speeds Divide nd Conquer Let's te cue from the movies! As you now, movie is just series of still imges, tht re run relly fst. Ech imge is still (no movement), ut series of them give the illusion of movement. Perhps we could te our velocity eqution, chop it up into series of still imges, nd get the motion from tht Let's sy it this wy: for ny short intervl of time, the chnge in velocity is not very lrge thus, for short intervl of time, we could consider the oject to e moving with constnt velocity, nd use our old fmilir distnce eqution. There is grphic interprettion of this ide creting rectngles round the grph of ( ) We'll strt y chopping the time intervl into four pieces: v t. Notice tht the oject is t rest round time.5 seconds thus, it turns round nd heds c in the other direction. Thus, the totl distnce trveled will e different from the net distnce trveled (displcement). Keep this in mind for lter! The totl re of these rectngles is 5 the totl distnce trveled is 5 meters. This is only n pproximtion, though we're supposed to e chopping the time intervl into sections where the velocity is lmost constnt! Let's do etter: try ten intervls. Tht gives totl re of This should e little etter however, there will still e some error! Let's try once gin, with intervls: HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

2 This gives n re of 4.8 this is etter, ut still not the nswer. Reductio d infinitum The nswer requires tht we consider n infinite numer of intervls! However, we'll leve tht for it, since tht requires limits. Let's te one step t time. Three Pths to the Summit The pictures tht I drew ove ll hve rectngles tht hve the left side on the function we cll these left-hnd rectngles. You could lso use midpoints (the height of the rectngle is t the midpoint of the top edge) or right sides (te wild guess!). You'll get slightly different nswers ut tht's oy; once we wor towrds the nswer, they'll ll come out the sme. Exmples [.] Estimte the re under the curve rectngles. y= e x in the intervl [, ] with four nd ten First, with four rectngles: The estimted re is.9. Now with ten rectngles: HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

3 The estimted re is.66. [.] The following dt were collected from moving vehicle: time (sec) velocity (ft/sec) Estimte the distnce trveled y the vehicle during the thirty second intervl. Note tht the dt hve lredy divided the intervl into prtition! The width of ech suintervl is 5 (seconds), nd the height of the rectngle is the given velocity (feet per second). width height re This gives totl re of 4 which mens totl distnce of 4 feet. 5.: Definite Integrls The Generic Pth: Riemnn Sums Let's generlize the ides we explored in the previous section A prtition of the intervl [, ] is set of vlues ( x i ) so tht < x < x <... < xn <. If we =, then P= { x, x, x... x n } is prtition of the intervl [, ]. cll x = nd xn The th suintervl of P is [ x x ], e the width of the th suintervl of P. x. Let c e ny vlue in the th suintervl of P. Also, let If f is function which is continuous on the intervl [, ], then the Riemnn sum of f (on tht intervl) is f ( c ) n = x. Note tht this is the sum of unch of rectngle res The Definite Integrl Bc when we were developing the ide of derivtives, we creted generic point, nd then moved the point ritrrily close to the trget point we llowed the distnce etween the points to pproch zero. This limiting process gve us the slope of the curve. Now consider the rectngles res from the Riemnn sum. One might thin tht dding more points to the prtition might me the rectngle res get relly smll, nd dd up to the re HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

4 under the curve however, just dding more points doesn't me ll of the rectngles smll. To do tht, you hve to dd points so tht the lrgest x pproches zero (the lrgest x is clled the norm of P, or P ). n If lim f ( c) x exists, then we sy tht f is integrle over the intervl [, ] P = vlue of the limit is clled the definite integrl of f over [, ] Nottion., nd the Tht limit nottion is terrile difficult to write, so we crete new nottion for the definite integrl of f over [, ] : ( ) f x dx. This is red s "the integrl of f from to." Evluting Definite Integrls Definite integrls re defined in such wy tht they pproximte the re etween curve nd the x-xis (if it lies ove the x-xis). However, since this is mth (nd not relity), the re etween the curve nd the x-xis for functions which lie elow the x-xis is negtive. Thus, the, is the re ove the xis plus the re elow the xis. definite integrl of f over [ ] If you relly wnt re (rel, positive re), then it's re ove the xis minus re elow the xis. Returning to our velocity exmple from while c: the totl distnce trveled will e the re etween the line v( t) = 9.8t+ 5 nd the x-xis. Fortuntely, this is just pir of tringles! First, note tht v (.55) =. The re of the left/upper tringle is ( 5 )(.55 ) =.888, nd the re of the right/lower tringle is ( 4.)( 4.55) =.88, for totl re of 4.76 ( totl distnce of 4.76 meters). The displcement is =.6 meters. Of course, rther thn doing ll of this y hnd, you could use your clcultor note tht the clcultor will do the definite integrl; not the re under the curve. My clcultor gives the vlue of the definite integrl t.6. Exmples 7 [.] Find dx. This is rectngle with se length of 8 nd height of ; the re (nd the vlue of the definite integrl) is 6. 5 [4.] Find ( ) 5 x dx. 5 =. This is tringle with se 5 nd height 5; the re is ( 5)( 5) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 4 OF

5 [5.] Find 4 x dx. This is semicircle of rdius. The re is 4π. 5.: Definite Integrls nd Antiderivtives Properties of the Definite Integrl Reversing the limits negtes the integrl: ( ) = ( ) f x dx f x dx The integrl of point is zero: f ( x ) dx= Constnts cn e fctored out: ( ) = ( ) f x dx f x dx Integrls cn e "distriuted" cross sums nd differences: ( ) ± ( ) = ( ) ± ( ) f x g x dx f x dx g x dx c c Integrls over djcent intervls cn e comined: ( ) + ( ) = ( ) f x dx f x dx f x dx The Averge Vlue of Function We ll now how to find n verge dd up ll the vlues, nd divide y the numer of vlues. It turns out tht you cn do tht with n infinite numer of things, lso! Adding n infinite numer of things cretes n integrl, though. The verge vlue of f ( x ) over the intervl [, ] is ( ) f x dx. The Men Vlue Theorem (for Definite Integrls) There is some point ( x= c ) in the intervl [, ] so tht f ( c) = f ( x) dx. The Men Vlue Theorem for Derivtives sys tht for ny intervl, there is some point where the slope of the tngent equls the slope of the secnt. The Men Vlue Theorem for Integrls sys tht for ny intervl, there is some point where the re of rectngle equls the re under the curve (( ) ( ) ( ) f c = f x dx ). The Connection: A First Glimpse The derivtive is found with quotient of infinitely smll differences. The definite integrl is found through sum of infinitely smll products (res). Sums nd differences quotients nd products surely there is some connection! nd there is. Here's loo hed which you'll need for few of the prolems in this section. ( ) = ( ) ( ), where F( x ) is ny ntiderivtive of ( ) f x dx F F f x. HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 5 OF

6 Exmples [6.] If f ( x) dx= 7 nd g( x) dx=6, then find ( ) + ( ) Using the properties of integrls, ( ) ( ) ( ) ( ) ( ) ( ) f x g x dx. f x + g x dx= f x dx+ g x dx= = [7.] Find ( ) x x dx. Since I don't now hndy formul for the re under prol, I'll use ntiderivtives. One f x = x x is F( x) = x x x. Thus, ( ) ( ) ( ) f x dx= F F F = + =, so ( ) 6 6 x x dx= = = ntiderivtive of ( ) F ( ) = = nd ( ) [8.] Find the verge vlue of on the intervl [, 4 ]. x 4 The formul gives me dx 4. I'll hve to use n ntiderivtive gin how out x x? 4 4 dx= x = = = x : The Fundmentl Theorem of Clculus FTC, Prt x d If g( x) = f ( t) dt, then g ( x ) f ( x ) dx =. In other words, the derivtive of n integrl is just the plin function. It's s if the derivtive nd the integrl re inverse opertions The hirs on the c of your nec should e stnding on end The Integrl s Function x The ide of g( x) ( ) = f t dt needs some dditionl investigtion there re lots of potentil questions tht cn e sed out functions tht re defined s integrls of other functions! Here re just few exmples from single scenrio The grph of f ( t ) is shown elow. It consists of semicircle nd line segment. HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 6 OF

7 = f t dt. x Define A( x) ( ) A ( 6) is the re under ( ) circle. ( ) ( ) π A = π = = 9π. 4 4 A ( 8) is the re under ( ) tringle! ( ) ( ) ( )( ) f t etween x= nd 6 f t etween x= nd 8 A 8 = π = 8π x= ( ) f t dt which is qurter HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 7 OF 8 x= : ( ) f t dt. This is semicircle nd Note tht A( x ) is n incresing function s x gets lrger, the mount of re under f ( t ) eeps getting lrger. = 8 =π+. 8 The verge vlue of f ( t ) on the intervl [,8 ] is f ( t) dt A( ) FTC, Prt 8 8 If f ( x ) is ny ntiderivtive of g( x ), then g ( x ) dx= f ( ) f ( ). Note tht f ( x ) eing n ntiderivtive of g( x ) mens tht f ( x) g( x) =. Thus, this theorem is ting the integrl of n ntiderivtive. Also note tht the integrl of this derivtive resulted in the originl function (sort of). Finding Are If you wnt to use definite integrls to find the re etween curve nd the x-xis, rememer tht res elow the xis will e negtive! One wy to find the totl re is to first determine where the function is equl to zero this will divide the res into ove/elow the x-xis. Add the integrls from the sections ove the xis, nd sutrct the integrls from the sections elow the xis. Another wy to do this is to just find the re under the solute vlue of the function. The solute vlue will simply force ll of the res to e positive. This mes things relly esy especilly if you re using your clcultor Exmples π cos x dx. [9.] Find ( ) Well n ntiderivtive of cos( x ) is ( ) π ( x ) dx ( x π ) ( ) ( ) etween sin( x ) nd the x-xis is zero! sin x so cos = sin ] = sin π sin = =. Note tht this does not men tht the re [.] Find the totl re etween f ( x) = x x nd the x-xis on the intervl [,] Let's e creful nd loo t the grph..

8 We'll need to split this up! The totl re will e ( ) ( ) ( ) ( ) x x dx+ x x dx x x dx+ x x dx 4 An ntiderivtive of f ( x) = x x is F( x) x x =. Tht gives us 4 F ( x ) F ( x ) F ( x ) F ( x ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) F( ) ( ) ( ) + ( ) ( ) + ( ) F F F F F Let's strt plugging in! F( ) = 6 4= 4 = 4 F( ) = = 4 4 F = ( ) F ( ) = = 4 4 F ( ) = 6 4= 4 = = = So finlly, we get ( ) d t dt dx. π tn x. x [.] Find tn( ) Esy! The derivtive is ( ) 5.5: The Trpezoidl Rule Another Pth Now tht we hve firm connection etween integrls nd re, let's pproch the re ide gin. Before, we used rectngles to pproximte the re under the curve simply ecuse HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 8 OF

9 rectngles re very simple geometric figures. Als, they don't relly do such good jo of pproximting re under the curve until you use lot of very tiny rectngles. There is nother shpe which does etter jo the trpezoid! Divide the intervl in question s you would efore, ut now connect the function vlues to otin trpezoids. For our liner exmple (wy c t the eginning of this chpter), we get n exct nswer with s few s two trpezoids! Let's te the function f ( x) = 4 ( x ) on the intervl [ ] exmple. Here is the pproximte re with four rectngles:, 4 with four regions s n (tht's n re of 9.848) Here is the pproximte re with four trpezoids: (tht's n re of 8.788) The ctul re is 9. You cn see tht the trpezoids do much etter jo of the pproximtion! Simpson's Rule/Approximtion You cn do even etter with your pproximtion if, insted of trpezoids, you use rectngles cpped with little prolic rches. It turns out tht there is (reltively simple) formul for this nd (firly simple) formul tht generlizes the entire procedure! It's clled Simpson's Rule: To pproximte f ( x ) dx, divide the intervl [, ] into n even numer of suintervls y o+ y + y + y + + y y y n + n + n. n Hppily, this topic is not prt of the AP Exm! (me n even) nd then clculte ( ) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE 9 OF

10 Exmples [.] Use four trpezoids to estimte the vlue of Let's loo t the picture: x dx. The res of the trpezoids re + = = = + = = = + = = = + 8 = = 8 8 This gives totl re of = 6 = 4.5. (the ctul re is 4) HOLLOMAN S AP CALCULUS AB ABC NOTES 5, PAGE OF

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