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# Chapter 5 Thomas 9 Ed F 2008

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### Chapter 5 Thomas 9 Ed F 2008

1. 1. CrnprpR Applications of Integrals OYERVIEW Many things we want to know can be calculated with integrals: the areasbetweencurves,the volumes and sudaceareasof solids, the lengthsof curves, the amount of work it takes to pump liquids fiom belowground, the forces againsr floodgates,the coordinatesof the points where solid objects will balance.We define all of these as limits of Riemann sums of continuous functions on closed intervals. that is, as integrals,and evaluatethese limits with calculus. There is a pattern to how we define the integrals in applications, a pattern that, once learned,enablesus to define new integralswhen we need them. We look at specific applications 0rst, then examine the pattern and show how it leads to integrals in new situations. AreasBetweenCurves This section shows how to lind the areas of regions in the coordinate plane bv integrating the functions that define the regions, boundaries. The Basic Formula a Limitof Riemann as Sums Supposewe want to find the area of a region that is bounded above by the curve -y =.l(r), below by the curve ) = g(.r), and on the left and right by the lines .r : 4 and r : D (Fig. 5.1). The region might accidentallyhave a shapewhose area w e c o u l d f i n L w i r h g e o m e t r l b u r i l / a n d g a r e i t r b i l r a r ) o n r i n u o ut,u n c r r o n,r l . c e usuaily have to flnd the area with an integral. To see what the integml should be, we lirst approximate the region with,? basedon a parlition p : {.r0, ... , r,,} of [ri, D] (Fig. 5.2, on verticalrectangles rr, the following page). The area of the tth rectangle(Fig. 5.3, on rhe followrng page; ls AA1 : tsjcfi1 x width : Ll (c.t) g(ci)l A4. We then approximatethe area of the region by adding the areasof the ,?rectanqles: A* f ^A, -Itr,,.r grr.,rlA,:,. ' The region between y: f(x) and As ll P | -+ 0 the sums on rhe right approachthe limit y: g ( x ) a n d t h e l i n e sx : a a n d x = b . _ g(,r)l d,r because /j t/(r) 365
2. 2. Chapter5: Applicationsof Integrals 356 T - s(ck) f(t) Stctll ::.: We approximate regionwith the perpendicular the x-axis. to AAk = areaof kth rectangle,f(c*) - g(cr) = height, lxk = width rectangles this integral' /and g arecontinuous. takethearea theregionto be thevalueof of We That is, lim )-l/(c,)-gtcrtlAxr - fo rf ,r,- A- Stxt]dx nP o-- J Definition with /(r) > g(x) throughout b], ihenthe area g [4, Ifl ancl arecontinuous the curves : /(*) andy : g(t) ftom a to b is the y of the regionbetween integral [/ - 8 | liom a to D: o[ rb U(x)- g(x)ldx. (1) A= J To apply Eq. (1) we take the following steps. How to Find the Area Between Two Curves Graph the cunes and draw a tepresentative rectangle. Thts rcveals l. which curve is / (upper curve) and which is g (1owercurye). It also helps find the limits of integration if you do not aheady know them. ,s'I ,=sinr 2. Find the limits of integration. I--''- 3. Write a formula for f(x) - g(x). Simplify it if you can. 4. lntegrate [f(x) - gQ, fron a to b Tl1e number you get is the area. H Find the area between lv = sec2 and y : 3i1;5 from 0 to z/4' x EXAMPLE 1 Solution Step t; We sketch the cutves and a vertical rectangle(Fig. 5.4). The upper curve is the gmph of /(;r) = secz;r: the lower is the graph of g(;) : sipI' Sfep 2; The limits of integration are already given: a :0' b = r /4' ,'.]].The regionin Example'lwith a - SteP3: f (x) - 8(x) = sec2;r sin't rectangle. typicalapproximating
3. 3. 5.1 AreasBetweenCurves 367 Step 4: {r' *- sinx)r/x : ftan + cos -r x]i/a lo'o f'.+l,to+:+ -t L f, Curves That Intersect When a region is detemined by curves that intersect, the intersection points give the limits of integration. - x2 Find the area of the region enclosedby the parabola ! :2 EXAMPLE 2 : -y. and the line y Solution Sfep t; Sketch the curves and a vertical rectangle (Fig. 5.5). Identifying the upper (-{,,f(-r) -x. The ,{-coordinates and the lower curyes, we take /(x) :2- g(r): xz arLd pointsare the limits of integration. of the intersecdon Step 2; We find the limits of integation by solving | :2 - xz and y : -1 ri- ( 1 ,1 ) multaneously for .r.' -x [ q r l L t/r1 r ) . u r d . q ( f r 2-xz : x2-x-2:0 R.rir.. (x+l)(x -2):0 Fxrr.' (a s(r)) -1, x: x :2. sot. The region runs from x : - 1 to.r : 2. The limits of integrationare a : - l, b : 2. Step 3: 5.5 The regionin Example with a 2 typicalapproximatingrectangle. Step 4 t2 f -2 -t12 - : o: rf,.,>e(x)ldx I t z + x - x z t d x = l z r + z - ,I l' JJ J-t I r / 4 8 / | 1 (o*r-r,l (-'*l*r/ ?qq - -'2 At 3 2 l Technofogy The Intersectionof Two Graphs One of the difficult and some- times frustrating parts of integmtion applications is finding the limits of inte- gration. To do this you often have to find the zeroes of a function or the intenection points of two curues. g(r) using a graphing utility, you enter To solve the equation /(x): yr: f@) and y2: g@)
4. 4. Chapter Applications lntegrals of 5: 368 you andusethe grapherroutine to find the Pointsof intersection'Altematively' - g(T) : 0 with a root finder'Try bothprocedures lun solve*tJ /(:r) quation with and g(x):3-x' f (x):lnx hidden When points of intersectionarc not clearly revealedor you suspect further use of calculus behavior,additional work with the graphing utility or may be necessary, ISECT =.79205996845 curves : Inx and lz:3 - x' usinga built-infundion y1 a) The intersecting intersection to find the - : x O) Usinga built-inroot finder to find the zero of f(x) Inx 3 + with ChangingFormulas Boundaries points' we partition If the formula for a bounding curve changesat one or mole th,gionintosubregionsthatconespondtotheformulachangesandapplyEq. (1) to eachsubregion. Find the areaof the region in the fint quadrantthat is bounded EXAMPLE 3 - 2' : aboveby y : .r/i and below by the t-axis and the line y x solution is the graphof Step t.' The sketch(Fig. 5.6) showsthat the region'supperboundary <2to g(x) = fromg(;r):0for0 S r changes f(; : JV. Theloweiboundary -' Z tot Z : t : 4 (thereis agreement r : 2)' We partition the region at r : 2 at rectanglefor eachsubregion' into subregiJns ani f and tketh a representative A 5.6 When the formulafor a boundingcurve changes, area integralchanges match to the (Example3). 'fe limits of integrationfor regionA area : O andb :,2' The lefchand step 2: li#t for region B rs a = 2. To find the right-hand limit, we solve the equations
5. 5. 5.1 AreasBetween Curves 369 y : rG andy : x - 2 simultaneously r.' for EqratcI (.r)rnd ^- - z v^ .9(11 a:(x.-2)2:x.2-4x+4 SqLrafeboth x 2- 5 x + 4 : o Re\$'r'ite. (.r-1)(r-4):0 FacL(r'. -_1 -_i Solve. Only the value.x=4 satisfies equation the The valuer:1is an Ji:x-2. extraneous introduced squaring. right-hand root by The limit is b:4. f (x)- S@): -0: S t e p3 . ' F o r 05 x = 2 : - e ( x ) : , 6E - ( x - 2q : ) Ji - x +2 F o r2 a x a 4 : f(x) Step 4.' We add the areaof subregions and B to find the total area: A f)-f4 = T o t a fr e a | J i d x I l t J x - x + 2 t d x a -o - - - - / J Jz areaof B arca of A 'l2 f) f1 -2 14 fil.+l1*'''-v+2,1, /2 - ?ef,,_o_ (1,,0,',, .' 8)_ _8 r r , r{zt'''-z++1 ?,t, : !. -, , 33 lntegrationwith Respect y to If a region's bounding curves are describedby functions of y, the approximating rectanglesare horizontal insteadof vertical and the basic formula has y in place of r. t 1 I L*,,, use tJteformula In Eq. (2),/always denotes right-hand the - stytldy. e: l.' vrrt (2) curve and I the left-hand curve, so fO) - e(y) is nonnegative.
6. 6. Chapter 5: Applicationsof lntegrals 370 Find the area of the region in Example 3 by integrating with EXAMPLE 4 rcspect to ). Solution (c()), ).) x:y+2 Step t; We sketchthe region and a typical horiz,ontalrectanglebasedon a panition of an interval ofy values(Fig. 5.7). The region's right-handboundary is the line x : y f 2, so l (y) : y + 2. The left-hand boundary is the curve ir : y2, so g(y) : y2. Step 2: The lower limit of integration is l = 0. We find the upper lirnit by solving I : y *2 and r: )2 simultaneously y; for 5.7 lt takestwo integrations find the l+t:l to areaof this regionif we integrate with 2=o ,t-) respectto x It takes only one if we with respect y (Example integrate to 4). ( ) + 1 ) ( ) 2 ): 0 Iurt,, 1'- _t ) S ,i l . The upperlimit of integmtion D : 2. (The valuey : -1 givesa point of inter- is sectlor belowthe r-axis.) Step3: =)+2 y2=2iy )? /())-s()) Step 4: v,rt- : e: 1,,'' srt,rat 1 2 + y - y ' ) l d y l,' f r,2 rt l' : +; 1l L2Y 3 ln 8 l0 - o.+ t - 4 3 3 This is the result of Example 3, fbund with less work. Integrals with Formulas from Geometry Combining andgeometry. The fastest way to find an areamay be to combinecalculus The Area of the Region in Example 3 Found the Fastesi EXAMPLE5 way Find the area of the region in Example 3. So/ution The area we want is the areabetweenthe curye I : Ji,O = x S 4, and 1:rf the x-axis, minus the area of a triangle with base 2 and height 2 (Fig. 5.8): 2 lr2rr2r / x r u = f or t a , .2 '.2 2 ,11 J .lo 5.8 The areaof the blue regionis the :3ttl-o -'z:+' area underthe parabola = 1& minus , y the areaof the triangle. JJ -
7. 7. Exercises 5.1 371 Moral of Examples3-5 It is sometimes easierto lind the areabetween two with respect y instead ,r. Also, it may help to combine curves integrating by to of geometry calculus. and After sketching region,takea moment determine the to the bestwav to oroceed. Exercises 5.1 Find the areasof the shadedregions in Exercises 1-8. l. 2.
8. 8. Chapter Applications Integrals of 5: 372 16. y : x 2 - 2 x and Y=r area. In Exercises findthetotalshaded 9-12, and y = -tt2 14x t7. ) : 12 10. 9, ] 18. y = l - l x ' ano ]:x-14 (-3,5) 19. y = x 4 - 4 x 2 + 4 and Y=a2 I and y:0 20, y = r n @ = 7 , a>0, _,_,2 5y = x * 6 (How many intersection points are 21. y = {fl nd there?) 2x3 - (x212)+4 5x 22,y:lx2-41'and ): 2 I Find the areasof the regions enclosedby the lines and curves 23-30. Exercises -3) : r : 0 , a n d) : 3 x:2y2, 23. 24. x = y2 u1t4 a: y 12 and' 4x - Y = 16 25. y2 - 4x:4 26, x - y2 :0 arld x +2Y2 =3 ^nd r+3Y2:2 2 7 ,x * y 2 : O and ,r*)a:2 2 8 .x - y 2 / t : 0 (-2,4 y 29.x =y2-l and x='yJl 3 0 . r = ) l 3- y 2 a n d x = 2 y -2 -l Find the areasof the regions enclosedby the curves in Exercises I 31'34. and .xa-Y:l 31.4x2+y:4 I a n d 3 x 2- y : 4 y :O 3 2 .x 3 I, (3,5) for xZ0 and :r*)a:1, 3 3 .x * 4 y 2 : 4 and 4.t*y2:0 3 4 ,x a y z : 3 12. Find the arcas of the regions enclosedby the lines and curves in ) I+. 35-42. Exercises (3,6) 6 and y = sin2.r, 0:.x57t 3 5 .y = 2 s 1 n x and y:5e92.r, -7t13=x Sr13 v 36.):8cos.r .t2 and 1:1 3 7 .y : c o s ( r x / 2 ) 38. y = sin(nx/2) and 1t:r (3,1) and'x:tt/4 39, l:sec2.r, y=tarfx, x:-tr/4, and .x:-tan2), -7r14=r 3n/4 4O. :ta#y x r and -x=0, 0=y<7t/2 4 1 .r : 3 s i n l / i o s y -' 4 2 .y : s s s 2 1 1 t a J 3 1 r r d y : r 1 / 1 , - l = i r : l 3) a rcgion enclosed the curve by 43. Find the areaof the propeller-shaped Find the areasof the regions enclosedby the lines and curves in - y3 : 0 andthe line.r - Y : 0. t 13-22. Exercises region enclosedby the 44. Find the area of the propeller-shaped 1 3 . y : a 2- 2 clruesx -yll3 : 0 andr - )l/5 = 0. ar'd y:2 14. y =2x - x2 arLd Y: -J 45, Find the areaof the region in the first quadrantboundedby the line I :1, the line,t :2, the curve1 : 1/-r'?, the -t-aris. and 15. y :1+ and ) :8r