APPLICATION OF INTEGRALS

APPLICATION OF INTEGRALS 59 Chpter 8 APPLICATION OF INTEGRALS One should study Mthemtics ecuse it is only through Mthemtics tht nture cn e conceived in hrmonious form. BIRKHOFF 8. Introduction In geometry, we hve lernt formule to clculte res of vrious geometricl figures including tringles, rectngles, trpezis nd circles. Such formule re fundmentl in the pplictions of mthemtics to mny rel life prolems. The formule of elementry geometry llow us to clculte res of mny simple figures. However, they re indequte for clculting the res enclosed y curves. For tht we shll need some concepts of Integrl Clculus. In the previous chpter, we hve studied to find the re ounded y the curve y f (x), the ordintes x, x nd x-xis, while clculting definite integrl s the limit of sum. Here, in this chpter, we shll study specific ppliction of integrls to find the re under simple curves, re etween lines nd rcs of circles, prols nd ellipses (stndrd forms only). We shll lso del with finding the re ounded y the ove sid curves. 8. Are under Simple Curves In the previous chpter, we hve studied definite integrl s the limit of sum nd how to evlute definite integrl using Fundmentl Theorem of Clculus. Now, we consider the esy nd intuitive wy of finding the re ounded y the curve y f (x), x-xis nd the ordintes x nd x. From Fig 8., we cn think of re under the curve s composed of lrge numer of very thin verticl strips. Consider n ritrry strip of height y nd width dx, then da (re of the elementry strip) ydx, where, y f (x). A.L. Cuchy (789-857) Fig 8.

6 MATHEMATICS This re is clled the elementry re which is locted t n ritrry position within the region which is specified y some vlue of x etween nd. We cn think of the totl re A of the region etween x-xis, ordintes x, x nd the curve y f (x) s the result of dding up the elementry res of thin strips cross the region PQRSP. Symoliclly, we express A da ydx f ( x) dx The re A of the region ounded y the curve x g (y), y-xis nd the lines y c, y d is given y d A xdy g( y) dy c c d Here, we consider horizontl strips s shown in the Fig 8. Fig 8. Remrk If the position of the curve under considertion is elow the x-xis, then since f (x) < from x to x, s shown in Fig 8., the re ounded y the curve, x-xis nd the ordintes x, x come out to e negtive. But, it is only the numericl vlue of the re which is tken into considertion. Thus, if the re is negtive, we tke its solute vlue, i.e., f ( x ) dx. Fig 8. Generlly, it my hppen tht some portion of the curve is ove x-xis nd some is elow the x-xis s shown in the Fig 8.. Here, A < nd A >. Therefore, the re A ounded y the curve y f (x), x-xis nd the ordintes x nd x is given y A A + A.

APPLICATION OF INTEGRALS 6 Exmple Find the re enclosed y the circle x + y. Solution From Fig 8.5, the whole re enclosed y the given circle (re of the region AOBA ounded y the curve, x-xis nd the ordintes x nd x ) [s the circle is symmetricl out oth x-xis nd y-xis] Fig 8. ydx (tking verticl strips) x dx Since x + y gives y ± x Fig 8.5 As the region AOBA lies in the first qudrnt, y is tken s positive. Integrting, we get the whole re enclosed y the given circle x x x + sin + sin π π

6 MATHEMATICS Alterntively, considering horizontl strips s shown in Fig 8.6, the whole re of the region enclosed y circle xdy y dy (Why?) y y y + sin + sin π π Fig 8.6 x y Exmple Find the re enclosed y the ellipse + Solution From Fig 8.7, the re of the region ABA B A ounded y the ellipse re of theregion AOBAin the first qudrnt ounded ythecurve, x xis nd theordintes x, x (s the ellipse is symmetricl out oth x-xis nd y-xis) ydx (tking verticlstrips) x y Now + gives y± x, ut s the region AOBA lies in the first qudrnt, y is tken s positive. So, the required re is x dx x x x + sin + sin (Why?) π π Fig 8.7

APPLICATION OF INTEGRALS 6 Alterntively, considering horizontl strips s shown in the Fig 8.8, the re of the ellipse is xdy y dy (Why?) y y y + sin + sin Fig 8.8 π π 8.. The re of the region ounded y curve nd line In this susection, we will find the re of the region ounded y line nd circle, line nd prol, line nd n ellipse. Equtions of ove mentioned curves will e in their stndrd forms only s the cses in other forms go eyond the scope of this textook. Exmple Find the re of the region ounded y the curve y x nd the line y. Solution Since the given curve represented y the eqution y x is prol symmetricl out y-xis only, therefore, from Fig 8.9, the required re of the region AOBA is given y xdy re of theregion BONBounded ycurve, y xis nd thelines y nd y Fig 8.9 ydy y 8 (Why?) Here, we hve tken horizontl strips s indicted in the Fig 8.9.

6 MATHEMATICS Alterntively, we my consider the verticl strips like PQ s shown in the Fig 8. to otin the re of the region AOBA. To this end, we solve the equtions x y nd y which gives x nd x. Thus, the region AOBA my e stted s the region ounded y the curve y x, y nd the ordintes x nd x. Therefore, the re of the region AOBA ydx [y ( y-coordinte of Q) (y-coordinte of P) x ] ( ) x dx (Why?) x 8 x Remrk From the ove exmples, it is inferred tht we cn consider either verticl strips or horizontl strips for clculting the re of the region. Henceforth, we shll consider either of these two, most preferly verticl strips. Exmple Find the re of the region in the first qudrnt enclosed y the x-xis, the line y x, nd the circle x + y. Y Solution The given equtions re y x... () y x nd x + y... () B (, ) Solving () nd (), we find tht the line nd the circle meet t B(, ) in the first qudrnt (Fig 8.). Drw perpendiculr A BM to the x-xis. X' X M O (, ) Therefore, the required re re of the region OBMO + re of the region BMAB. Now, the re of the region OBMO ydx xdx... () x 8 Fig 8. Y' Fig 8.

APPLICATION OF INTEGRALS 65 Agin, the re of the region BMAB ydx x dx x x x + sin + sin 6 + sin 8 π (8 + π) π 8... () Adding () nd (), we get, the required re π. Exmple 5 Find the re ounded y the ellipse nd x e, where, ( e ) nd e <. x y + nd the ordintes x Solution The required re (Fig 8.) of the region BOB RFSB is enclosed y the ellipse nd the lines x nd x e. Y Note tht the re of the region BOB RFSB B S x e e ydx e x dx F( e, o) X X e O x x x + sin R B' e e sin e + Y Fig 8. e e + sin e EXERCISE 8.. Find the re of the region ounded y the curve y x nd the lines x, x nd the x-xis.. Find the re of the region ounded y y 9x, x, x nd the x-xis in the first qudrnt.

66 MATHEMATICS. Find the re of the region ounded y x y, y, y nd the y-xis in the first qudrnt.. Find the re of the region ounded y the ellipse x y +. 6 9 5. Find the re of the region ounded y the ellipse x y +. 9 6. Find the re of the region in the first qudrnt enclosed y x-xis, line x y nd the circle x + y. 7. Find the re of the smller prt of the circle x + y cut off y the line x. 8. The re etween x y nd x is divided into two equl prts y the line x, find the vlue of. 9. Find the re of the region ounded y the prol y x nd y x.. Find the re ounded y the curve x y nd the line x y.. Find the re of the region ounded y the curve y x nd the line x. Choose the correct nswer in the following Exercises nd.. Are lying in the first qudrnt nd ounded y the circle x + y nd the lines x nd x is π π π (A) π (B) (C) (D). Are of the region ounded y the curve y x, y-xis nd the line y is (A) (B) 9 8. Are etween Two Curves Intuitively, true in the sense of Leinitz, integrtion is the ct of clculting the re y cutting the region into lrge numer of smll strips of elementry re nd then dding up these elementry res. Suppose we re given two curves represented y y f (x), y g (x), where f (x) g(x) in [, ] s shown in Fig 8.. Here the points of intersection of these two curves re given y x nd x otined y tking common vlues of y from the given eqution of two curves. For setting up formul for the integrl, it is convenient to tke elementry re in the form of verticl strips. As indicted in the Fig 8., elementry strip hs height (C) 9 (D) 9

APPLICATION OF INTEGRALS 67 f(x) g(x) nd width dx so tht the elementry re Fig 8. da [f (x) g(x)] dx, nd the totl re A cn e tken s A [() f x g ()] x dx Alterntively, A [re ounded y y f (x), x-xis nd the lines x, x ] [re ounded y y g (x), x-xis nd the lines x, x ] [ ] f ( xdx ) gxdx ( ) f ( x) g( x) dx, where f (x) g (x) in [, ] If f (x) g (x) in [, c] nd f (x) g (x) in [c, ], where < c < s shown in the Fig 8., then the re of the regions ounded y curves cn e written s Totl Are Are of the region ACBDA + Are of the region BPRQB c [ ( ) ( )] + [ ( ) ( )] f x g x dx g x f x dx c Y A C y f( x) B y g ( x) P R D y g ( x) x x c Q y f( x) x X O Y Fig 8. X

68 MATHEMATICS Exmple 6 Find the re of the region ounded y the two prols y x nd y x. Solution The point of intersection of these two prols re O (, ) nd A (, ) s shown in the Fig 8.5. Here, we cn set y x or y x f(x) nd y x g (x), where, f (x) g (x) in [, ]. Therefore, the required re of the shded region [ ( ) ( )] f x g x dx x x x dx x P (, ) Thus, the points of intersection of these two curves re O(, ) nd P(,) ove the x-xis. From the Fig 8.6, the required re of the region OPQCO included etween these two curves ove x-xis is Y Fig 8.6 (re of the region OCPO) + (re of the region PCQP) 8 ydx + ydx Exmple 7 Find the re lying ove x-xis nd included etween the circle x + y 8x nd the prol y x. Solution The given eqution of the circle x + y 8x cn e expressed s (x ) + y 6. Thus, the centre of the Y circle is (, ) nd rdius is. Its intersection with the prol y x gives x + x 8x or x x or x (x ) X X or x, x O C (, ) 8 x dx + x dx (Why?) ( ) Fig 8.5 Q (8, )

APPLICATION OF INTEGRALS 69, where, x + t dt x t t t + t sin + (Why?) + sin + π + 8 + + π (8+ π) Exmple 8 In Fig 8.7, AOBA is the prt of the ellipse 9x + y 6 in the first qudrnt such tht OA nd OB 6. Find the re etween the rc AB nd the chord AB. Solution Given eqution of the ellipse 9x + y 6 cn e expressed s x y + or 6 y x + nd hence, its shpe is s given in Fig 8.7. 6 Accordingly, the eqution of the chord AB is y 6 ( ) x or y (x ) or y x + 6 Are of the shded region s shown in the Fig 8.7. (Why?) x dx (6 x) dx x sin x 6 x x + x Fig 8.7 + sin () π 6 π 6

7 MATHEMATICS Exmple 9 Using integrtion find the re of region ounded y the tringle whose vertices re (, ), (, ) nd (, ). Solution Let A(, ), B(, ) nd C (, ) e the vertices of tringle ABC (Fig 8.8). Are of ΔABC Are of ΔABD + Are of trpezium BDEC Are of ΔAEC Now eqution of the sides AB, BC nd CA re given y y (x ), y x, y Hence, re of Δ ABC (x ), respectively. x x dx x dx dx ( ) + ( ) x x x x + x x + Exmple Find the re of the region enclosed etween the two circles: x + y nd (x ) + y. Solution Equtions of the given circles re x + y... () nd (x ) + y... () Eqution () is circle with centre O t the origin nd rdius. Eqution () is circle with centre C (, ) nd rdius. Solving equtions () nd (), we hve (x ) + y x + y or x x + + y x + y or x which gives y ± Thus, the points of intersection of the given circles re A(, ) nd A (, ) s shown in the Fig 8.9. Fig 8.8 Fig 8.9

APPLICATION OF INTEGRALS 7 Required re of the enclosed region OACA O etween circles [re of the region ODCAO] (Why?) [re of the region ODAO + re of the region DCAD] y dx + y dx ( x ) dx + x dx (Why?) x ( x ) ( x ) + sin + x x x + sin x x ( x ) ( x ) + sin + x x + sin + sin sin ( ) + sin sin π π π π + + 6 6 π π + π + π 8 π EXERCISE 8.. Find the re of the circle x + y 9 which is interior to the prol x y.. Find the re ounded y curves (x ) + y nd x + y.. Find the re of the region ounded y the curves y x +, y x, x nd x.. Using integrtion find the re of region ounded y the tringle whose vertices re (, ), (, ) nd (, ). 5. Using integrtion find the re of the tringulr region whose sides hve the equtions y x +, y x + nd x.

7 MATHEMATICS Choose the correct nswer in the following exercises 6 nd 7. 6. Smller re enclosed y the circle x + y nd the line x + y is (A) (π ) (B) π (C) π (D) (π + ) 7. Are lying etween the curves y x nd y x is (A) (B) (C) Miscellneous Exmples Exmple Find the re of the prol y x ounded y its ltus rectum. Solution From Fig 8., the vertex of the prol y x is t origin (, ). The eqution of the ltus rectum LSL is x. Also, prol is symmetricl out the x-xis. The required re of the region OLL O (re of the region OLSO) ydx xdx x dx X O Y (D) L (,) S X x Y Fig 8. L' 8 8 Exmple Find the re of the region ounded y the line y x +, the x-xis nd the ordintes x nd x. Solution As shown in the Fig 8., the line y x + meets x-xis t x nd its grph lies elow x-xis for x, nd ove x-xis for x,. Fig 8.

APPLICATION OF INTEGRALS 7 The required re Are of the region ACBA + Are of the region ADEA (x + ) dx + (x + ) dx x x + x + + x 5 + 6 6 Exmple Find the re ounded y the curve y cos x etween x nd x π. Solution From the Fig 8., the required re re of the region OABO + re of the region BCDB + re of the region DEFD. Thus, we hve the required re Fig 8. π π cos xdx + cos xdx + cos xdx π π π π π [ sin x] + [ sin x] + [ sin x] + + π π π Exmple Prove tht the curves y x nd x y divide the re of the squre ounded y x, x, y nd y into three equl prts. Solution Note tht the point of intersection of the prols y x nd x y re (, ) nd (, ) s Fig 8.

7 MATHEMATICS shown in the Fig 8.. Now, the re of the region OAQBO ounded y curves y x nd x y. x x dx x x 6 6... () Agin, the re of the region OPQAO ounded y the curves x y, x, x nd x-xis x dx x 6... () Similrly, the re of the region OBQRO ounded y the curve y x, y-xis, y nd y y xdy dy y 6... () From (), () nd (), it is concluded tht the re of the region OAQBO re of the region OPQAO re of the region OBQRO, i.e., re ounded y prols y x nd x y divides the re of the squre in three equl prts. Exmple Find the re of the region {(x, y) : y x +, y x +, x } Solution Let us first sketch the region whose re is to e found out. This region is the intersection of the following regions. A {(x, y) : y x + }, A {(x, y) : y x + } nd A {(x, y) : x } X' O Y Y' Q(, ) R x P (,) x Fig 8. The points of intersection of y x + nd y x + re points P(, ) nd Q(, ). From the Fig 8., the required region is the shded region OPQRSTO whose re re of the region OTQPO + re of the region TSRQT T S X ( x + ) dx + ( x + ) dx (Why?)

APPLICATION OF INTEGRALS 75 x x + x + + x ( ) + + + + 6 Miscellneous Exercise on Chpter 8. Find the re under the given curves nd given lines: (i) y x, x, x nd x-xis (ii) y x, x, x 5 nd x-xis. Find the re etween the curves y x nd y x.. Find the re of the region lying in the first qudrnt nd ounded y y x, x, y nd y.. Sketch the grph of y x + nd evlute x + dx. 6 5. Find the re ounded y the curve y sin x etween x nd x π. 6. Find the re enclosed etween the prol y x nd the line y mx. 7. Find the re enclosed y the prol y x nd the line y x +. 8. Find the re of the smller region ounded y the ellipse x y line +. 9. Find the re of the smller region ounded y the ellipse x y line +. x y + nd the 9 x y + nd the. Find the re of the region enclosed y the prol x y, the line y x + nd the x-xis.. Using the method of integrtion find the re ounded y the curve x + y. [Hint: The required region is ounded y lines x + y, x y, x + y nd x y ].

76 MATHEMATICS. Find the re ounded y curves {(x, y) : y x nd y x }.. Using the method of integrtion find the re of the tringle ABC, coordintes of whose vertices re A(, ), B (, 5) nd C (6, ).. Using the method of integrtion find the re of the region ounded y lines: x + y, x y 6 nd x y + 5 5. Find the re of the region {(x, y) : y x, x + y 9} Choose the correct nswer in the following Exercises from 6 to. 6. Are ounded y the curve y x, the x-xis nd the ordintes x nd x is (A) 9 (B) 5 7. The re ounded y the curve y x x, x-xis nd the ordintes x nd x is given y (A) (B) (C) (C) [Hint : y x if x > nd y x if x < ]. 8. The re of the circle x + y 6 exterior to the prol y 6x is 5 (D) (D) 7 (A) ( π ) (B) ( π+ ) (C) (8 π ) (D) (8 π+ ) π 9. The re ounded y the y-xis, y cos x nd y sin x when x is (A) ( ) (B) (C) + (D) Summry The re of the region ounded y the curve y f (x), x-xis nd the lines x nd x ( > ) is given y the formul: Are ydx f ( x) dx. The re of the region ounded y the curve x φ (y), y-xis nd the lines y c, y d is given y the formul: Are xdy φ( y) dy d c d c.

APPLICATION OF INTEGRALS 77 The re of the region enclosed etween two curves y f (x), y g (x) nd the lines x, x is given y the formul, [ ] Are f ( x) g( x) dx, where, f (x) g (x) in [, ] If f (x) g (x) in [, c] nd f (x) g (x) in [c, ], < c <, then c [ f x g x ] dx [ g x f x ] dx. Are ( ) ( ) + ( ) ( ) c Historicl Note The origin of the Integrl Clculus goes ck to the erly period of development of Mthemtics nd it is relted to the method of exhustion developed y the mthemticins of ncient Greece. This method rose in the solution of prolems on clculting res of plne figures, surfce res nd volumes of solid odies etc. In this sense, the method of exhustion cn e regrded s n erly method of integrtion. The gretest development of method of exhustion in the erly period ws otined in the works of Eudoxus ( B.C.) nd Archimedes ( B.C.) Systemtic pproch to the theory of Clculus egn in the 7th century. In 665, Newton egn his work on the Clculus descried y him s the theory of fluxions nd used his theory in finding the tngent nd rdius of curvture t ny point on curve. Newton introduced the sic notion of inverse function clled the nti derivtive (indefinite integrl) or the inverse method of tngents. During 68-86, Leinitz pulished n rticle in the Act Eruditorum which he clled Clculs summtorius, since it ws connected with the summtion of numer of infinitely smll res, whose sum, he indicted y the symol. In 696, he followed suggestion mde y J. Bernoulli nd chnged this rticle to Clculus integrli. This corresponded to Newton s inverse method of tngents. Both Newton nd Leinitz dopted quite independent lines of pproch which ws rdiclly different. However, respective theories ccomplished results tht were prcticlly identicl. Leinitz used the notion of definite integrl nd wht is quite certin is tht he first clerly pprecited tie up etween the ntiderivtive nd the definite integrl. Conclusively, the fundmentl concepts nd theory of Integrl Clculus nd primrily its reltionships with Differentil Clculus were developed in the work of P.de Fermt, I. Newton nd G. Leinitz t the end of 7th century.

78 MATHEMATICS However, this justifiction y the concept of limit ws only developed in the works of A.L. Cuchy in the erly 9th century. Lstly, it is worth mentioning the following quottion y Lie Sophie s: It my e sid tht the conceptions of differentil quotient nd integrl which in their origin certinly go ck to Archimedes were introduced in Science y the investigtions of Kepler, Descrtes, Cvlieri, Fermt nd Wllis... The discovery tht differentition nd integrtion re inverse opertions elongs to Newton nd Leinitz.