Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

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1 Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9

2

3 Introduction to Ares Integrtion Sttement of Prerequisite Skills Complete ll previous TLM modules efore eginning this module. Required Supporting Mterils Access to the World Wide We. Internet Eplorer 5.5 or greter. Mcromedi Flsh Pler. Rtionle Wh is it importnt for ou to lern this mteril? Lerning Outcome When ou complete this module ou will e le to Lerning Ojectives. Find the re etween curve nd coordinte is.. Find the re s sum of seprte portions.. Find the re etween two curves. Connection Activit Module A Ares Integrtion

4 OBJECTIVE ONE When ou complete this ojective ou will e le to Find the re etween curve nd coordinte is. Eplortion Activit Are To determine the ect re under the curve in Figure elow, divide the line segment AB into numer of segments, s, of uniform width Δ. The division points re,,, 4 nd 5, with nd corresponding to points A nd B respectivel. Now divide the desired re into strips drwing verticl lines through the points on AB nd intersecting the curve. The totl re under the curve is ectl equl to the sum of the res of these strips. = f() O Δ 4 5 A B Figure To estimte the res of the strips we form perfect rectngles s re shown in Figure nd cll these rectngles I, II, III, IV, V nd VI. The comined re of these rectngles is less tht the totl re under the curve. Module A Ares Integrtion

5 = f() I II III IV V VI O Δ 4 5 A B Figure The sum of the res of the rectngles cn e epressed s: f ( ) Δ + f( ) Δ + f( ) Δ + f( ) Δ + f( ) Δ + f( ) Δ 4 5 Using summtion nottion: Are of rectngles = f ( ) Δ 5 i= 5 i= nd f( ) Δ < totl re under the curve i i Now, if the re under the curve is divided into strips insted of, nd rectngles re formed, the re of the rectngles is still slightl less thn the ect re under the curve. However, the comined re of the rectngles will e more nerl equl to the ect re under the curve. Module A Ares Integrtion

6 If the numer of strips increses without limit so tht the width Δ of the strips pproches zero, the comined re of ll the rectngles pproches the ect re under the curve s limit. Tht is: n lim f( i ) Δ = ect re under the curve n = i The fundmentl theorem of integrl clculus sttes tht the limit of sum of this kind is equl to the definite integrl etween n two given vlues of the independent vrile. n B lim f ( i ) Δ = f( ) d n A = i or more riefl stted we hve, B Are = f ( ) d () A Similrl the re etween curve nd the is, with limits on going from P to Q, cn e found using the formul: Are = Conclusion: = Q P Q P d f ( ) d(). When ou perform definite integrtion ou re finding n re under curve.. The definite integrl provides mens of solving re prolems. 4 Module A Ares Integrtion

7 EXAMPLE Find the re etween the stright line = nd the is from = to = 8. SOLUTION: Using formul with f( ) =, = nd = 8, we get: A= 8 = d 8 8 = 4 9 = 55 = squre units. The shded re in grph elow illustrtes this. = O 8 Module A Ares Integrtion 5

8 EXAMPLE Find the re etween the curve = nd the is from = to = 5. SOLUTION: This is the sme grph s in Emple, ut the re is ounded the is. Using formul () with f ( ) =, p = nd q = 5, we get: A= = 5 d 5 ( ) ( ) 5 = 75 = =. squre units. This is illustrted the shded re in following grph. is nd not the 5 4 = O 5 5 Module A Ares Integrtion

9 Eperientil Activit One. Find the re etween the is nd the stright line 5= + 9 from = to = 4.. Determine the re etween the is nd the stright line = 5+ from to = 9. =. Determine the re ounded the is nd the stright line 9= 4 from = to = Determine the re ounded the is nd the stright line 8 = from to = 5. = 5. Find the re etween the is nd the stright line = 5 from = to = 4.. Determine the re ounded the is nd the stright line 7= 4 from = 4 to =. Eperientil Activit One Answers Module A Ares Integrtion 7

10 EXAMPLE Find the re ounded =, the is, =, nd =. SOLUTION: Using formul from Ojective with f ( ) =, = nd =, we get; A= = = d d = 8 = = 9. squre units. The shded re in the following grph illustrtes this: 4 = O 4 8 Module A Ares Integrtion

11 EXAMPLE 4 Determine the re ounded =, =, =, nd =. 5. SOLUTION: This emple cn est e solved if we first mke sketch of the required re..5. =.5 O 4 8 The re is ounded the is, therefore we should use horizontl rectngles nd tke our limits from the is, s goes from to.5. A= c 5. d d 5. = d= =. 75 squre units. Note in this emple tht the limits were not eplicitl stted, ut were otined from the digrm nd the sttement of the prolem. Module A Ares Integrtion 9

12 EXAMPLE 5 Find the re ounded the curve (NOTE: This emple is optionl.) + 4 7=, from =, to =. SOLUTION: A sketch is necessr to get n ide of wht re we re to find. M = From the sketch we see the is of smmetr of the prol is =. NOTE: The theor for finding re under curve hs one ver ig limittion: The rectngle strip cnnot strt nd stop on the sme curve. So wht we do is find the re etween the curve nd the line =, nd then simpl doule this vlue to recognize the other hlf of the totl re we re to find. Thus the height of our rectngle in the shded re is ( ) = +. So our shded re is: A= ( + ) d N = ([(4 + 8) ] + ) d = (4+ 8) ( ) = 4 = 4.44 d ( ) To see how we sustitute with (4 +8) / see the notes on the net pge. Douling this we get the totl re to e 8.88 squre units. Module A Ares Integrtion

13 Notes: for the previous process: To get in terms of, solve + 4 7= for completing the squre. + = = ( ) + = = 4+ 8 = 4+ 8 ( ) / = This emple illustrtes fct to consider: Finding the re under curve is ver strightforwrd ut ou must e creful in defining the length of our rectngle. The in the formul A= d m e inomil, or some other lgeric epression. EXAMPLE Find the re in the first qudrnt ounded = 9 d O d (,) SOLUTION A: The digrm ove shows the re to e found. Our solution involves integrting with respect to the is. The limits of integrtion must e long the is. The lower limit is t =. A little lger is necessr to find the upper limit ( intercept where = ): Module A Ares Integrtion

14 9 = t the intercept = 9 =± from = 9 We ignore = (not in first qudrnt) The upper limit of is =. Now we find the re with respect to the is: A= f( ) d (9 ) = = 9 d () = 9() = (7 9) = 8. squre units. SOLUTION B: The question in Emple 4 could lso e done integrting long the The limits of integrtion must e long the is. The lower limits long is =. is. A little lger is necessr to find the upper limits ( intercept where = ): from = 9 = 9 = 9 The upper limit long is = 9 Now we find the re with respect to the is. Solve the originl eqution for : = 9 = 9 = (9 ) Sustitute this epression for in the following formul: Module A Ares Integrtion

15 A= d 9 = 9 d 9 = (9 ) ( d) (9 ) = + = (9 ) = (9 9) + (9 ) = + (7) = 8. squre units sme re s Solution A REMEMBER: In d, it is the d which indictes wht letter we re integrting with respect to nd thus which letter the function eing integrted must e epressed in s n independent vrile. Thus must e replced with n epression in efore integrtion tkes plce. Similrl, in d, the d tells us the independent vrile is nd the must e replced with n epression in (solve the function for ). Module A Ares Integrtion

16 Eperientil Activit One (Continued). Determine the re etween the is nd the curve = 7 + 7from = to = 5. Find the re etween the is nd = + + from =. to.. Determine the re etween the is nd the curve = 7 from = to = 4 4. Determine the re etween the is nd the curve = 4to=. 5. Determine the re etween the is nd the curve = 4 8 from = = = from to 7. Find the re in the first qudrnt ounded the is nd the curve = =. =. 5 from 7 to 5 Eperientil Activit One Answers (Continued) / / Module A Ares Integrtion

17 OBJECTIVE TWO When ou complete this ojective ou will e le to Find the re s sum of seprte portions. Eplortion Activit Sometimes the re we re to find is not completel on the sme side of n is, tht is, prt of it m e on one side, nd nother prt on the other side of the is. In cses such s this ou simpl find the re of ech prt nd then dd the solute vlues of ech prt. Rememer, re is sclr quntit, not vector quntit, therefore there is no mening to negtive vlue for re. A sketch of the curve helps to solve this tpe of prolem. EXAMPLE Determine the totl re ounded the is nd the line + =, s goes from to 8. SOLUTION: I + = Step : Mke digrm 4 II Step : The res we wnt re I nd II Step : The line crosses the is t = Step 4: Integrte from to nd gin from to 8 nd dd the solute vlues of ech integrl to otin: 8 A= ( + ) d + ( + ) d Where stnd for solute vlue A = + A = 5 squre units Module A Ares Integrtion 5

18 Eperientil Activit Three. Find the totl re ounded the is nd the stright line + 8 = from = 4 to =. Find the totl re ounded the is nd the stright line = 4 from = to = 5. Determine the totl re ounded the is nd the stright line + 5= 4 from = to = 7 4. Find the totl re ounded the is nd the stright line = + from = 4to =5 5. Determine the totl re ounded the is nd the stright line 7 = from = to = 7. Determine the totl re ounded the is nd the stright line + 8 = 8 from = 4 to = 4 Eperientil Activit Three Answers / /4. 74/ Module A Ares Integrtion

19 OBJECTIVE THREE When ou complete this ojective ou will e le to Find the re etween two curves. Eplortion Activit Are Between Two Curves So fr in this module we hve used the definite integrl is used to compute the re of region ounded the function = f() nd coordinte is. In this section we will use the definite integrl to compute res of more generl regions. Specificll, let us now consider regions tht re ounded oth ove nd elow grphs of functions. Refer to following figure. = f() = g() O d In the ove grph the upper curve is = f() nd the lower curve is = g(). Prolem Let f() nd g() e functions tht re continuous for ll etween nd, where <. We wish to determine simple epression for the re of the shded region under the grph of = f() nd ove the grph of = g() from = to =. Module A Ares Integrtion 7

20 Oservtion: The region referred to is the shded region in the Figure. We see tht the region is ounded on the left the verticl line = nd on the right the verticl line =. We shll refer to the re of this region simpl s the re etween the curves nd or the re etween the curves f() nd g() from = to =. (The nming f() s nd g() s is ritrr, s eplined lter.) SOLUTION: This prolem, too hs rther strightforwrd solution. It is the region under the grph of f ( ) minus the region under the grph of g( ) : A= f( ) d g( ) d A= [ f( ) g( )] d or A= ( ) d Solving for A gives us the re etween curve nd curve. Are etween two curves: If = f() lies ove = g() from = to =, the re of the region etween f() nd g() from = to = is: A= [ f( ) g( )] d or A= ( ) d Remrks: ( ). We m choose & ritrril using A=. If & intersect t point c, nd < c <, then c ( ) ( ) A = d + c d d 8 Module A Ares Integrtion

21 EXAMPLE Find the re etween = nd = +. Illustrte sketching grph. SOLUTION: The following grph illustrtes the solution: = + = O Let = nd = +. The choice for nd is ritrr, the solute vlue of the nswer comes out the sme. Solve nd for their common points of intersection, thus ielding the integrtion limits, letting = : ( )( ) = = + = + = = or + = = = Our limits of integrtion ecome: = (lowerlimit) = (upper limit) The re of the shded region, from formul, ecomes: Module A Ares Integrtion 9

22 A= ( ) d ( ) A= + d ( ) A= + + d A= A = A = = 9 A = squre units EXAMPLE Find the re etween 4 = + nd = +. Illustrte mking sketch. SOLUTION: The following sketch illustrtes the solution: = +4 = O Let = +4 nd = +. Module A Ares Integrtion

23 Solving simultneousl for the limits of integrtion: = + 4= + = (+ )( ) = = = The limits of integrtion re = nd =. The re ecomes: A= ( ) d = [( + ) ( + 4)] A= ( + 4) d A= ( + + ) d A d A= A = A = 54 Module A Ares Integrtion

24 EXAMPLE Find the re etween the prol = nd the stright line + =. SOLUTION: Solving the two equtions simultneousl the points of intersection re (.7,.79) nd (.7,.). : = 4 : + = 4 O 4 Using verticl strips we integrte from =.7 to =.7. The length of ech strip is nd the width is d, therefore we get: 7. A= ( ) d A= d 7. + A= A = Note: In generl when using verticl strips we sutrct the lower from the upper. When using horizontl strips we sutrct the left from the right. d The formul is: A= ( ) d. See the net emple. c Module A Ares Integrtion

25 EXAMPLE 4 Find the re etween = nd =. Use horizontl elements. SOLUTION: The student m wish to mke sketch, however digrm isn t solutel necessr. = is prol with its verte t the origin nd opening to the right. = is stright line tht psses through the origin nd hs slope =. Using d c A= ( ) d, with = d A= ( ) d c, nd = we hve: =, from c to d. () 4 We still must solve for the limits c nd d. Equte nd : = = = 5 = ( 5) = = nd5 Sustituting these limits into () ove nd we hve: A = A = 4 = 5 = 5 5 A = A = 5 5 A = = tking the solute vlue we get A = 5 squre units. Module A Ares Integrtion

26 Eperientil Activit Four. Find the re etween the stright line = 7+ 8 nd the curve = + 7. Determine the re etween the stright line = nd the prol =. Determine the re etween the curves = + nd = + 4. Find the re etween the curves = nd = 5. Determine the re etween the curves 4 7 = + + nd = Determine the re etween the curves (see shpe s in emple ) = + 5 nd = Find the re etween the curves from =. to =. 4 5 = + + nd the curve 4 5 = Find the re ounded the curves = + + nd = Find the re ounded the curves = + nd = Eperientil Activit Four Answers. 4/ /. / / 9. / 4 Module A Ares Integrtion

27 Prcticl Appliction Activit Complete the Ares Integrtion ssignment in TLM. Summr This module demonstrted how integrls cn e used to determine res etween curves. Module A Ares Integrtion 5

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