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 After completing this Section ou should be ble to... understnd integrtion s the reverse of differentition be ble to use tble of integrls 3 be ble to evlute definite integrls 4 be ble to sketch grphs of common functions including polnomils, simple rtionl functions, eponentil nd trigonometric functions find the re bounded b curve nd the -is find the re between two curves

2 . Clculting the Are under Curve Let us denote the re under = f() between fied point nd vrible point b A(): = f() re is A() It 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 the figure below. = f() f() re is A( +δ) A( ) + δ To good pproimtion the chnge in the re is: A( + δ) A() f()δ n pproimtion which gets better nd better s δ gets smller nd smller, nd so: f() Clerl, in the limit s δ 0wehve A( + δ) A() δ A( + δ) A() f() = lim δ 0 δ But this limit on the right-hnd side is the derivtive of A() with respect to so f() = da d Thus A() isnindefinite 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 A(b) A() [A()] b = 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 (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve

3 . The Are Bounded b Curve Ling Above the -is Consider the grph of the function = f() shown in the figure below. 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 Ke Point 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 re = 4 d = [ln ]4 =ln4 ln = ln 4 =.386 (3dp) 3 HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve

4 Find the re bounded b the curve = sin nd the -is between =0 nd = π. (The required re is shown in the figure. Note tht it lies entirel bove the -is. =sin re required O π Your solution π sin d =[ cos 0 ]π 0 =. Find the re under f() =e from =to =3given tht the eponentil function e is lws positive. Your solution (Note: Becuse e is positive, the re will lie bove the -is.) re = 3 e d = [ e] 3 = 98 HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve 4

5 Emple The figure shows the grphs of = sin nd = cos for 0 π. The two grphs intersect t the point where = π. Find the shded re. 4 =cos re required =sin π Solution To find the shded re we could clculte the re under the grph of = sin for between 0 nd π, nd subtrct this from the re under the grph of = cos between the sme 4 limits. Alterntivel the two processes cn be combined into one nd we cn write shded re = π/4 0 (cos sin )d = [sin + cos ] π/4 0 = ( sin 4 π + cos 4 π) (sin 0 + cos 0) If ou re wre of the stndrd tringles ou will know tht sin 4 π = cos 4 π = in which cse the vlue of integrl is =0.44. Alterntivel ou cn use our clcultor to obtin this result directl. Eercises In ech question the required re lies entirel bove the horizontl is, lthough ou should verif this fct for ourself independentl.. 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 =0nd =. 3. Find the re bounded b the curve =3t nd the t-is between t = 3 nd t =3. 4. Find the re under = between =nd =0.. 73,. 4, 3. 54, HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve

6 3. The Are Bounded b Curve, Prts of Which Lie Below the -is The figure below shows grph of = +. = re required 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 + ) ( ) ( = 7 3 += 4 3 The evlution of the re ields negtive quntit. There is 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 If n re contins prts both bove nd below the horizontl is, cre must be tken when tring to clculte 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. Emple Find the totl re enclosed b the curve = nd the -is between =0nd =3. HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve 6

7 Solution We need to determine which prts of the grph, if n, 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 =0, =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. If ou hve ccess to grphics clcultor or computer pckge this is trivil mtter. The grph is shown below. = re required From the grph we see tht the required re lies prtl bove the -is (when 0 ) nd prtl below (when 3). So we evlute the integrl in two prts: Firstl: 0 [ ( )d = ] 0 = ( 4 5 ) 3 + (0) = 7 This is the prt of the required re which lies bove the -is. Secondl: 3 ( )d = = [ 4 ] ( ) 3 +8 ( 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 7 HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve

8 ) Sketch the grph of = sin for 0 π. b) Find the totl re bounded b the curve nd the -is between = 3 π nd = 3 4 π. Your solution Sketch the grph nd indicte the required re noting where the grph crosses the -is. Perform the integrtion in two prts to obtin the required re. π/ π/3 sin d = 4 nd 3π/4 π/ sin d =. The required re is 4 + = 3 4. Eercises. Find the totl re enclosed between the -is nd the curve = 3 between = nd =.. Find the re under = cos t from t =0tot = Find the re enclosed b =4 nd the is from ) =0to =,b) = to =,c) =to =3. 4. 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. Answers ) 6 3,b)9,c) e. 6. HELM (VERSION : Mrch 8, 004): Workbook Level 3.3: The Are Bounded b Curve 8