13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes

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1 The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce trvelled b vehicle. In this Section we eplin how such n re is clculted. Prerequisites Before strting this Section ou should... Lerning Outcomes On completion ou should be ble to... understnd integrtion s the reverse of differentition be ble to use tble of integrls be ble to evlute definite integrls be ble to sketch grphs of common functions including polnomils, simple rtionl functions, eponentil functions nd trigonometric functions find the re bounded b curve nd the -is find the re between two curves 4 HELM (8): Workbook 3: Integrtion

2 . Clculting the re under curve Let us denote the re under = f() between fied point nd vrible point b A(): re is A() = f() Figure 6 A() is clerl function of since s the upper limit chnges so does the re. How does the re chnge if we chnge the upper limit b ver smll mount δ? See Figure 7 below. = f() f() re is A( +δ) A( ) + δ Figure 7 To good pproimtion the chnge in the re is: A( + δ) A() f()δ [This is becuse the shded re is pproimtel rectngle with bse δ nd height f().] This pproimtion gets better nd better s δ gets smller nd smller. Rerrnging gives: A( + δ) A() f() δ Clerl, in the limit s δ we hve f() = lim δ A( + δ) A() δ But this limit on the right-hnd side is the derivtive of A() with respect to so f() = da() d Thus A() is n indefinite integrl of f() nd we cn therefore write: A() = f()d Now the re under the curve from to b is clerl A(b) A(). But remembering our shorthnd nottion for this difference, introduced in the lst Section we hve, finll [ ] b b A(b) A() A() = f()d We conclude tht the re under the curve = f() from to b is given b the definite integrl of f() from to b. HELM (8): Section 3.3: The Are Bounded b Curve 5

3 . The re bounded b curve ling bove the -is Consider the grph of the function = f() shown in Figure 8. Suppose we re interested in clculting the re underneth the grph nd bove the -is, between the points where = nd = b. When such n re lies entirel bove the -is, s is clerl the cse here, this re is given b the definite integrl b f() d. re required = f() b Figure 8 Ke Point 4 The re under the curve = f(), between = nd = b is given b when the curve lies entirel bove the -is between nd b. b f() d Emple Clculte the re bounded = nd the -is, between = nd = 4. Solution Below is grph of =. The re required is shded; it lies entirel bove the -is. = re required O Figure 9 re = 4 [ ] 4 d = ln = ln 4 ln = ln 4 =.386 (3 d.p.) 6 HELM (8): Workbook 3: Integrtion

4 Tsk Find the re bounded b the curve = sin nd the -is between = nd =. (The required re is shown in the figure. Note tht it lies entirel bove the -is.) =sin re required O Your solution [ sin d = ] cos =. Tsk Find the re under f() = e from = to = 3 given tht the eponentil function e is lws positive. Your solution re = 3 e d = [ ] 3 e = 98 to 3 significnt figures. HELM (8): Section 3.3: The Are Bounded b Curve 7

5 Emple 3 The figure shows the grphs of = sin nd = cos for. The two grphs intersect t the point where =. Find the shded re. 4 =cos re required =sin Figure Solution To find the shded re we could clculte the re under the grph of = sin for between nd, nd subtrct this from the re under the grph of = cos between the sme limits. 4 Alterntivel the two processes cn be combined into one nd we cn write shded re = /4 (cos sin )d [ ] /4 = sin + cos = ( sin + cos ) (sin + cos ) 4 4 = ( + ) ( + ) = = So the numeric vlue of the integrl is =.44 to 3 d.p.. (Alterntivel ou cn use our clcultor to obtin this result directl b evluting sin 4 nd cos 4.) Eercises In ech question ou should check tht the required re lies entirel bove the horizontl is.. Find the re under the curve = 7 nd bove the -is between = nd = 5.. Find the re bounded b the curve = 3 nd the -is between = nd =. 3. Find the re bounded b the curve = 3t nd the t-is between t = 3 nd t = Find the re under = between = nd =.. 73,. 4, 3. 54, HELM (8): Workbook 3: Integrtion

6 3. The re bounded b curve, not entirel bove the -is Figure shows grph of = +. = re required Figure The shded re is bounded b the -is nd the curve, but lies entirel below the -is. Let us evlute the integrl ( + )d. ( + )d = = [ 3 ] 3 + ) ( ) ( = = 4 3 The evlution of the re ields negtive quntit. There is, of course, no such thing s negtive re. The re is ctull 4, nd the negtive sign is n indiction tht the re lies below the -is. 3 (However, in pplictions of integrtion such s work/energ or distnce trvelled in given direction negtive vlues cn be meningful.) If n re contins prts both bove nd below the horizontl is, cre must be tken when clculting this re. It is necessr to determine which prts of the grph lie bove the horizontl is nd which lie below. Seprte integrls need to be clculted for ech piece of the grph. This ide is illustrted in the net Emple. HELM (8): Section 3.3: The Are Bounded b Curve 9

7 Emple 4 Find the totl re enclosed b the curve = nd the -is between = nd = 3. Solution We need to determine which prts of the grph lie bove nd which lie below the -is. To do this it is helpful to consider where the grph cuts the -is. So we consider the function nd look for its zeros = ( 5 + 4) = ( )( 4) So the grph cuts the -is when =, = nd = 4. Also, when is lrge nd positive, is lrge nd positive since the term involving 3 domintes. When is lrge nd negtive, is lrge nd negtive for the sme reson. With this informtion we cn sketch grph showing the required re: 3 4 = re required Figure From the grph we see tht the required re lies prtl bove the -is (when ) nd prtl below (when 3). So we evlute the integrl in two prts: Firstl: [ ] ( ( + 4)d = = 4 5 ) 3 + () = 7 This is the prt of the required re which lies bove the -is. Secondl: 3 ( )d = = [ 4 ] ( ) ( 4 5 ) 3 + = 3 This represents the prt of the required re which lies below the -is. The ctul re is. 3 Combining the results of the two seprte clcultions we cn find the totl re bounded b the curve: re = = 95 3 HELM (8): Workbook 3: Integrtion

8 Tsk () Sketch the grph of = sin for. (b) Find the totl re bounded b the curve nd the -is between = 3 nd = 3 4. () Sketch the grph nd indicte the required re noting where the grph crosses the -is: Your solution =sin (b) Perform the integrtion in two prts to obtin the required re: Your solution / sin d = 3/4 4 nd sin d =. /3 / The required re is 4 + = 3 4. HELM (8): Section 3.3: The Are Bounded b Curve 3

9 Eercises. Find the totl re enclosed between the -is nd the curve = 3 between = nd =.. Find the re under = cos t from t = to t = Find the re enclosed b = 4 nd the is () from = to =, (b) from = to =, (c) from = to = Clculte the re enclosed b the curve = 3 nd the line =. 5. Find the re bounded b = e, the -is nd the line =. 6. Find the re enclosed between = ( )( ) nd the is. s () 6 3, (b) 9, (c) e 6. 3 HELM (8): Workbook 3: Integrtion