Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions of Are............................................. 5 2 The Definite Integrl 7 2. VIDEO - Where Are We Going With This?...................................... 7 2.2 VIDEO - Sigm Nottion................................................ 8 2.3 VIDEO - The Definite Integrl............................................. 0 3 The Fundmentl Theorem of Clculus 2 3. VIDEO - Prt of the FToC.............................................. 2 3.2 VIDEO - Prt 2 of the FToC.............................................. 4 4 Indefinite Integrls nd Net Chnge 6 4. VIDEO - Indefinite Integrls.............................................. 6 4.2 VIDEO - Net Chnge Theorem............................................. 7 5 The Substitution Rule 8 5. VIDEO - Indefinitely................................................... 8 5.2 VIDEO - Definitely.................................................... 9 5.3 VIDEO - Additionl Properties............................................. 20 Chpter 5 2 Are Between Curves 2. VIDEO - X Direction.................................................. 2.2 VIDEO - Y Direction.................................................. 23 5 Averge Vlue of Function 24 5. VIDEO - We ll love good verge.......................................... 24 Pge
Motivtion to Chpters 4 & 5 In Chpters 4 nd 5 we will bring up the second min topic: re under (the grph of) function. We will see tht surprisingly this is relted to derivtives. We will lern how to find the re under more complicted functions nd the re trpped between two functions. But why do we cre? Well s usul it s becuse there re rel world pplictions. Remrk 0.. If f(x) is rte of chnge then the re under f(x) on the intervl [, b] is the totl chnge of from x = to x = b. Exmple 0.2. A pump is delivering wter into tnk t rte of r(t) = + t 2 liters/minute, where t is time in minutes since the pump ws turned on. How much wter hs been pumped in the first 6 minutes? r(t) 6 t tlk bout units And so nytime you hve rte of chnge nd you re interested in totl chnge you will wnt re under functions. You hve velocity nd wnt displcement You hve rte of chnge of n infection nd wnt the totl number of people infected You hve mrginl profit nd wnt totl profit. Pge 2
Ares nd Distnces. VIDEO - Ares Under Functions Estimte the re under curve using rectngles with heights given by left endpoints or right endpoints. Solve for over/under estimtes for the re under curve using rectngles. Consider the following problem which helps to emphsize some issues Exmple.. Find re between f(x) nd the x xis on the intervl [, 4]. 4 f(x) 3 2 0 x 0 2 3 4 Exmple.2. Approximte the re between f(x) nd the x xis on the intervl [, 4] using 3 rectngles of equl width. The gol of this section is to understnd how to pproximte res with rectngles. Suppose tht we wnt to pproximte the re between the grph of continuous function f(x) nd the x-xis between x = nd x = b (suppose for now tht f is positive). Let s sub-divide the intervl [, b] into n sub-intervls [x 0, x ] through [x n, x n ] of equl width x. If we pick point x i in ech intervl [x i, x i+ ], then we cn estimte the re under the grph by the sum of the res of the rectngles with width x nd height f(x i ): Are f(x 0) x + f(x ) x + + f(x n ) x b Pge 3
But the question remins of wht x i re the should we pick? Definition(s).3. An upper sum is when the x i re ll chosen to be the globl mximum on [x i, x i+ ] for ech i. A lower sum is when the x i re ll chosen to be the globl minimum on [x i, x i+ ] for ech i. A left-hnd sum is when x i is the left endpoint of [x i, x i+ ] for ech i. A right-hnd sum is when x i is the right endpoint of [x i, x i+ ] for ech i. Exmple.4. Consider the continuous function g(x) whose vlues re given in the tble below x 0 0.5.0.5 2.0 g(x) 6 4 2 5 Assume ll bsolute mximums nd minimums of g(x) re contined in the bove points. () Using 4 rectngles find the upper sum of g(x) on the intervl [0, 2]. (b) Using rectngles of width 0.5 find the lower sum of g(x) on the intervl [0, 2]. (c) Using x = 0.5 find the left-hnd sum of g(x) on the intervl [0, 2]. (d) Using n = 4 find the right-hnd sum of g(x) on the intervl [0, 2]. Pge 4
.2 VIDEO - Applictions of Are Argue tht re under curve of velocity function will yield displcement. Solve vrious ppliction problems. You my hve lerned in physics clss tht if you hve constnt rte (think velocity) then distnce = rte time Exmple.5. Suppose runner trvels t 20 feet per second. How fr does she trvel between t = seconds nd t = 9 seconds? ply up units now do grph And so we see tht ctully tht the distnce (or more ccurtely displcement) is the re under the velocity grph. And in this wy we cn extend this ide to more relistic situtions. Exmple.6. Suppose runner s velocity (in ft/s) is given by the curve to the right. Sketch the re tht represents how fr she trveled between t = seconds nd t = 9 seconds. 9 8 v(t) 7 6 5 4 3 2 2 3 4 5 6 7 8 9 t Exmple.7. Approximte the distnce the runner trveled between t = nd t = 9 seconds using left-hnd sum nd 4 rectngles. Sketch this re on the grph bove Pge 5
Exmple.8. Suppose the velocity of runner is given by v(t) = t meters per second. Approximte the distnce the runner trveled from t = 0 to t = 4 seconds. () Using time intervls of second nd left-hnd sum (b) Using time intervls of second nd right-hnd sum Pge 6
2 The Definite Integrl 2. VIDEO - Where Are We Going With This? Lern how to get better pproximtions for re under curve. Setup why we wnt to be ble to dd together 00,,000, or,000,000 numbers For bit now we hve been pproximting the re under function using res of rectngles. Now we fce the issue of ccurcy. Minly, if you wnt to be s ccurte s possible how mny rectngles should you use? The nswer... with more rectngles comes more ccurcy. Check out https://www.desmos.com/clcultor/cxsfmpvf69 to see this in ction. Oky so we wnt more rectngles... so wht? Well let s try problem to see wht chllenges we will fce. Exmple 2.. Approximte the re under f(x) = x on the intervl [, 2] using right-hnd sums nd 00 rectngles. x = /00 0 00 00 + 02 00 00 + 03 00 00 + 04 00 (0 + 02 + 03 + 04 + + 200) 0000 00 + + 200 00 00 Pge 7
2.2 VIDEO - Sigm Nottion Express sums using sigm nottion. Memorize few common finite sums. Understnd bsic properties of finite sums nd use them to compute more complicted finite sums. Definition(s) 2.2. If m, m+,..., n, n re rel numbers nd m nd n re integers such tht m n, then i = m + m+ + + n + n i=m Definition(s) 2.3. This wy of short-hnding sums of mny numbers is clled sigm nottion the Greek letter Σ Sigm ). The letter i bove is clled the index of summtion (uses nd it tkes on consecutive integer vlues strting with m nd ending with n. Theorem 2.4. If c is ny constnt then: () (b) (c) c i = c i=m i=m i ( i + b i ) = i + i=m i=m i=m ( i b i ) = i i=m i=m i=m b i b i Theorem 2.5. Let c be constnt nd n positive integer. Then () (b) (c) = n i= i = i= i 2 = i= n(n + ) 2 n(n + )(2n + ) 6 Pge 8
Exmple 2.6. Write the sum: 3 + 4 + + 25 in sigm nottion Exmple 2.7. Evlute the following sums () 4 (2 3i) i= (b) 400 (2 3i) i= Pge 9
2.3 VIDEO - The Definite Integrl Use the limit of finite sums to clculte the definite integrl of function. Identify how the definite integrl reltes with re under the curve. If more rectngles re good then why not hve 000, 00000, or idelly even infinitely mny rectngles. Then the re would be so super ccurte tht there wouldn t be ny difference t ll! Exmple 2.8. Using 8 rectngles nd right-hnd sum pproximte the re under the curve f(x) = x on [, 4]. Do not simplify nything. Now to mke the jump to infinitely mny rectngles let s strt off by clculting the re under n rectngles. Exmple 2.9. Using n rectngles nd right-hnd sum pproximte the re under the curve f(x) = x on [, 4]. Do not simplify nything. Pge 0
Remrk 2.0. () The width of rectngle is given by x = b n (b) The right-hnd endpoint of the i th rectngle is given by x i = + i x Definition(s) 2. (Definite Integrl). If f is continuous on [, b], then f(x) dx = lim n i= f(x i ) x Remrk 2.2. The definite integrl, f(x) dx gives the net re Tht is, if the function is below the x xis the re is counted negtively. Exmple 2.3. Evlute the definite integrl 7 between the curve f nd the x xis. f(x) dx. Where f(x) is given by the function to the right. Exmple 2.4. Use the definition of the definite integrl to evlute 2 x dx. y x Pge
3 The Fundmentl Theorem of Clculus 3. VIDEO - Prt of the FToC Stte the first prt of the FToC. Go over n ide of the proof to see why it is probbly true. Apply the FToC, properties of definite integrls, nd the chin rule to tke some derivtives. Exmple 3.. A pump is delivering wter into tnk t rte of r(t) = + t 2 since the pump ws turned on. Wht does w(x) = x 0 r(t) dt represent? Evlute w() nd w(3). liters/minute, where t is time in minutes r(t) wht re the units? 6 t Theorem 3.2 (FTC, Prt ). If f is continuous on [, b], then the function g defined by g(x) = x f(t) dt, x b is continuous on [, b] nd differentible on (, b), nd g (x) = f(x) Remrk 3.3. Here is n ide of the proof: Pge 2
Exmple 3.4. Find the derivtive of g(x) = x t 2 dt Exmple 3.5. Find the derivtive of f(x) = 2 x t2 + 3 dt x 3 Exmple 3.6. Find the derivtive of H(x) = sin(2t 2 ) dt Pge 3
3.2 VIDEO - Prt 2 of the FToC Stte the second prt of the FToC. Go over n ide of the proof to see why it is probbly true. Use the ntiderivtive to clculte definite integrls. Theorem 3.7 (FTC, Prt 2). If f is continuous on [, b], then f(x) dx = F (b) F () where F is ny ntiderivtive of f, tht is, function such tht F = f. Remrk 3.8. Here is n ide of the proof: Remrk 3.9. The two prts of the FTC together stte tht differentition nd integrtion re inverse processes. Remrk 3.0. From our perspective FTC, Prt 2 is the most importnt becuse it llows us to clculte definite integrls without hving to tke limits of Riemnn sums! Pge 4
Nottion 3.. F (b) F () will lso be denoted F (x) b or [ F (x) ] b. Exmple 3.2. Evlute the following integrls () 9 x dx (b) 5 (u + )(u 2) du (c) 2 x 4 + x 2 dx Pge 5
4 Indefinite Integrls nd Net Chnge 4. VIDEO - Indefinite Integrls Define Indefinite integrls nd introduce some of their properties. Prctice computing indefinite integrls. Definition(s) 4.. An indefinite integrl of f, written f(x) dx, is n ntiderivtive of f. In other words, F (x) = f(x) dx mens F (x) = f(x) Theorem 4.2 (Linerity Properties of Indefinite Integrls). For functions f(x) nd g(x), nd ny constnt k R, [f(x) + g(x)] dx = f(x)dx + g(x)dx nd kf(x) dx = k f(x)dx Exmple 4.3. Find the most generl function F (x) with the property tht: () F (x) = x + cos(x) (b) F (x) = sec(x) (tn(x) + sec(x)) Pge 6
4.2 VIDEO - Net Chnge Theorem Use the Net Chnge Theorem to continue clculting definite integrls. Given velocity function nd set time intervl find the totl distnce trveled. Theorem 4.4 (Net Chnge Theorem). The net chnge of quntity F (x) from x = to x = b is the integrl of its derivtive from to b. Tht is, if F (x) = f(x), then F (b) F () = F (x) dx = f(x) dx In other words, F (b) = F () + F (x) dx = F () + f(x) dx Remember tht definite integrls re sums nd the derivtive is the instntneous rte of chnge. So the Net Chnge Theorem sys if you sum up the instntneous rtes of chnge you get totl chnge. And when you put it like tht it sounds lmost obvious. Remrk 4.5. If s(t) represents the position of moving object, nd v(t) the velocity, then the Net Chnge Theorem sys we cn compute the net displcement of the object from time t = to time t = b by integrting the velocity: s (t) dt = v(t) dt = s(b) s() Exmple 4.6. An object moves long line with velocity v(t) = 8t t 2 inches per second. At time t = 0, the object is 5 inches to the right of the origin. Wht is the position of the object t time t = 6 seconds? Pge 7
5 The Substitution Rule 5. VIDEO - Indefinitely Develop substitution rule to find ntiderivtives of more complicted functions. Apply it to couple of problems. Theorem 5.. If u = g(x) is differentible function whose rnge is n intervl I nd f is continuous on I, then f(g(x)) g (x) dx = f(u) du Remrk 5.2. The Substitution Rule sys essentilly tht is is permissible to operte with dx nd du fter the integrl signs s if they were differentils. Exmple 5.3. Evlute the following indefinite integrls: () x 2 x 3 + dx (b) dt ( 3t) 5 Pge 8
5.2 VIDEO - Definitely Upgrde our use of the substitution rule to include definite integrls. Apply the substitution rule to more problems. Theorem 5.4. If g (x) is continuous on [, b] nd f is continuous on the rnge of u = g(x), then f(g(x))g (x)dx = g(b) g() f(u)du Remrk 5.5. This chnge in the limits of integrtion is nnoying nd somewht unnecessry right now (so long s you chnge your vrible bck) however in Clc II once you strt doing trig substitutions it becomes extremely useful. WeBWorK will force you to prctice chnging the limits of integrtion so we will prctice here too. Exmple 5.6. Evlute the following definite integrls by chnging the limits of integrtion ppropritely: () 0 3 + 7x dx (b) π 0 x cos(x 2 ) dx Pge 9
5.3 VIDEO - Additionl Properties Introduce properties of definite integrls nd symmetric functions to crete few more theorems. Use them! Definition(s) 5.7. Recll gin () f is clled even if f( x) = f(x). (b) f is clled odd if f( x) = f(x). Theorem 5.8. Suppose f is continuous on [, ] then: () If f is even, then (b) If f is odd, then f(x) dx = 2 0 f(x) dx = 0 f(x) dx Exmple 5.9. Evlute π/4 π/4 tn 2 (x) sec 2 (x) dx Exmple 5.0. Evlute π/4 π/4 ( + x + x 2 tn x) dx Pge 20
Are Between Curves. VIDEO - X Direction Express the re bounded by two curves s definite integrl nd evlute. Del with more difficult situtions like when grphs cross. Theorem.. If f(x) nd g(x) re continuous functions on [, b] where f(x) g(x) for ll x in [, b], then the re of the region in between the grphs of y = f(x) nd y = g(x) between x = nd x = b is given by Are = (f(x) g(x)) dx Exmple.2. Find the re bounded between y = nd y = 5 x 2. y x Remrk.3. This theorem only pplies if f(x) g(x) on [, b]. In the more generl cse (where the grphs cross), we cn use the following theorem. Theorem.4. If f(x) nd g(x) re continuous functions on [, b], then the re of the region in between the grphs of y = f(x) nd y = g(x) between x = nd x = b is given by Are = f(x) g(x) dx Pge 2
Exmple.5. Find the re between y = x 2 nd y = 3 2x on the intervl [0, 3]. Hint: Here s wht this sitution looks like https://www.desmos.com/clcultor/tytqdpfng2 One more thing tht sometimes comes up is tht the function cn switch on us. Exmple.6. Sketch the lines y = x, y = 4 x nd y = 4 x 3. Setup definite integrl or integrls tht represents the re bounded between them 4 3 2 2 3 4 x Pge 22
.2 VIDEO - Y Direction Identify when it is dvntgeous to integrte with respect to y insted of x. Get bit more prctice. Remrk.7. If it is more convenient, you cn think of x s function of y, nd integrte with respect to y. For exmple, to find the re between the grphs of x = f(y) nd x = g(y) between y = nd y = b, compute Are = f(y) g(y) dy Exmple.8. Find the re enclosed by the line y = x nd the prbol y 2 = 2x + 6. Tke look t this region vi: https://www.desmos.com/clcultor/ynjlvejtqf Pge 23
5 Averge Vlue of Function 5. VIDEO - We ll love good verge Visulize the verge vlue of function on n intervl. Clculte the verge vlue of function over n intervl. Relte the verge vlue of function to verges of finite sets. This is how I visulize the verge vlue of function b x b x b x Definition(s) 5.. The verge vlue of function f(x) on the intervl [, b] is defined to be: f ve = b f(x)dx Pge 24
Exmple 5.2. Find the verge vlue of the following functions on the given intervl: () f(x) = x on [0, 4] (b) f(x) = sin(x) on [0, π] Definition(s) 5.3. Given set of n vlues {y, y 2,..., y n } the verge vlue, or men, of the set is denoted y nd is given by: y = y + y 2 +... + y n n Remrk 5.4. Brek up [, b] into n subintervls nd see how the verge of f over finite set is relted to our definition. Pge 25