Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus (vi) f ( ) d f ( ) d if f ( ) f ( ) if f ( ) f (

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1 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Clculus 7 DEFINITE INTEGRALS In the previous lesson we hve discussed the nti-derivtive, i.e., integrtion of function.the very word integrtion mens to hve some sort of summtion or comining of results. Now the question rises : Why do we study this rnch of Mthemtics? tics? In fct the integrtion helps to find the res under vrious lmins when we hve definite limits of it. Further we will see tht this rnch finds pplictions in vriety of other prolems in Sttistics, Physics, Biology, Commerce nd mny more. In this lesson, we will define nd interpret definite integrls geometriclly, evlute definite integrls using properties nd pply definite integrls to find re of ounded region. OBJECTIVES After studying this lesson, you will e le to : define nd interpret geometriclly ly the definite integrl s limit of sum; evlute given definite integrl using ove definition; stte fundmentl theorem em of integrl clculus; stte nd use the following properties for evluting definite integrls : c c ( ) ) (i) f d f ( d ( ) ( ) ( ) (ii) f d f d + f d (iii) (iv) (v) f ( ) d f ( ) d + f ( ) d f ( ) d f ( + ) d f ( ) d f ( ) d ww.pic ching g.com MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

2 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus (vi) f ( ) d f ( ) d if f ( ) f ( ) if f ( ) f ( ) (vii) f ( ) d f( ) d if f is n even function of if f is n odd function of. pply definite integrls to find the re of ounded region. Definite Integrls EXPECTED BACKGROUND KNOWLEDGE Knowledge of integrtion Are of ounded region 7. DEFINITE INTEGRAL AS A LIMIT OF SUM In this section we shll discuss the prolem of finding the res of regions whose oundry is not fmilir to us. (See Fig. 7.) Fig. 7. Fig. 7. Let us restrict our ttention to finding the res of such regions where the oundry is not fmilir to us is on one side of -is only s in Fig. 7.. finw ching ing.c om This is ecuse we epect tht it is possile to divide ny region into few suregions of this kind, find the res of these suregions nd finlly dd up ll these res to get the re of the whole region. (See Fig. 7.) Now, let f () e continuous function defined on the closed intervl [, ]. For the present, ssume tht ll the vlues tken y the function re non-negtive, so tht the grph of the function is curve ove the -is (See. Fig.7.). MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

3 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Clculus Fig. 7. Consider the region etween this curve, the -is nd the ordintes nd, tht is, the shded region in Fig.7.. Now the prolem is to find the re of the shded region. In order to solve this prolem, we consider three specil cses of f () s rectngulr region, tringulr region nd trpezoidl region. The re of these regions se verge height In generl for ny function f () on [, ] Are of the ounded region (shded region in Fig. 7. ) se verge height The se is the length of the domin intervl [, ]. The height t ny point is the vlue of f () t tht point. Therefore, the verge height is the verge of the vlues tken y f in [, ]. (This my not e so esy to find ecuse the height my not vry uniformly.) Our prolem is how to find the verge vlue of f in [,]. 7.. Averge Vlue of Function in n Intervl If there re only finite numer of vlues of f in [,], we cn esily get the verge vlue y the formul. Sumof.pf pt thevluesof fin Averge vlue of f in [,] [, ] Numersof vlues But in our prolem, there re infinite numer of vlues tken y f in [, ]. How to find the verge in such cse? The ove formul does not help us, so we resort to estimte the verge vlue of f in the following wy: First Estimte : Tke the vlue of f t '' only. The vlue of f t is f (). We tke this vlue, nmely f (), s rough estimte of the verge vlue of f in [,]. Averge vlue of f in [, ] ( first estimte ) f () (i) Second Estimte : Divide [, ] into two equl prts or su-intervls. ww.p.com Let the length of ech su-intervl e h, h. Tke the vlues of f t the left end points of the su-intervls. The vlues re f () nd f ( + h) MATHEMATICS 5 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

4 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus (Fig. 7.) Definite Integrls Fig. 7. Tke the verge of these two vlues s the verge of f in [, ]. Averge vlue of f in [, ] (Second estimte) f ( ) + f ( + h), h (ii) This estimte is epected to e etter estimte e thn the first. Proceeding in similr mnner, divide the intervl [, ] into n suintervls of length h (Fig. 7.5), h n Fig. 7.5 Tke the vlues of f t the left end points of the n suintervls. The vlues re f (), f ( + h),...,f [ + (n-) h]. Tke the verge of these n vlues of f in [, ]. Averge vlue of f in [, ] (nth estimte) ]ẇ ching.c ( ) + ( + ) + + ( ( + ) ) f f h... f n h, n h (iii) n For lrger vlues of n, (iii) is epected to e etter estimte of wht we seek s the verge vlue of f in [, ] Thus, we get the following sequence of estimtes for the verge vlue of f in [, ]: 6 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

5 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls f () [ f ( ) + f ( + h ) ], h [ f ( ) + f ( + h ) + f ( + h )], h [ f ( ) + f ( + h ) f{ + ( n ) h} ], h n n As we go frther nd frther long this sequence, we re going closer nd closer to our destintion, nmely, the verge vlue tken y f in [, ]. Therefore, it is resonle e to tke the limit of these estimtes s the verge vlue tken y f in [, ]. In other words, Averge vlue of f in [, ] lim { f ( ) + f ( n n + h ) + f ( + h ) +... f+ [ ( +n+n ) h ]}, h n (iv) It cn e proved tht this limit eists for ll continuous us functions f on closed intervl [, ]. Now, we hve the formul to find the re of the shded region in Fig. 7., The se is ( ) nd the verge height is given y (iv). The re of the region ounded y the curve f (), -is, the ordintes nd ( ) lim { f ( ) + f ( + h ) + f+f km ( + h )... + f + [ ( n+ ) h ]}, n n lim [ f ( ) + f ( + h ) f{ + ( n ) h} ],h n n n (v) We tke the epression on R.H.S. of (v) s the definition of definite integrl. This integrl is denoted y ww.pf{ chi com ( ) f d Clculus red s integrl of f () from to '. The numers nd in the symol ( ) f d re clled respectively the lower nd upper limits of integrtion, nd f () is clled the integrnd. Note : In otining the estimtes of the verge vlues of f in [, ], we hve tken the left end points of the suintervls. Why left end points? Why not right end points of the suintervls? We cn s well tke the right end points of the MATHEMATICS 7 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

6 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus suintervls throughout nd in tht cse we get f d lim f h f h... f, ( ) ( ) { ( + ) + ( + ) + + ( )} h n n [ ( ) ( ) ( ) ] Definite Integrls h n lim h f + h + f + h f (vi) Emple 7. Find Solution : By definition, d s the limit of sum. f d lim f f h... f n h, [ ] ( ) ( ) ( ) + ( + ) { ( + ) } h n Here,, f () nd h n n ( ). n n d lim f... f n n +f f + n n n lim n n + + n n n yc n lim M n km n ntimes n n n km km lim n n n n n ( ( ) ) ( ) ch cn n.n lim n n n + n. chi om ( ) n.n Since ( n ) n lim n n lim n n 8 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

7 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Emple 7. Find Solutions : By definition e d s limit of sum. [ ] ( ) ( ) + ( + ) + ( + ) + + { ( + ) } f d lim h f f h f h... f n h where h h n Here,,f ( ) e nd h ed lim h[ f ( ) + f ( h ) + f ( h ) +... f+ ( n ) h ] h h h ( n ) h lim h e + e + e... + e+ h lim h e h h ( ) n n e h e MATHEMATICS 9 n n r Since + r + r r r nh h e lim h lim h h h h e h e h e e ych My e ( nh ) h e e lim h eh h eh lim e h h In emples 7. nd 7. we oserve tht finding the definite integrl s the limit of sum is quite difficult. In order to overcome this difficulty we hve the fundmentl theorem of integrl clculus which sttes tht Theorem : If f is continuous in [, ] nd F is n ntiderivtive of f in [, ] then ( ) ( ) ( ) f d F F The difference F () F () is commonly denoted y [ F( ) ] om ( ) ( )] [ ( )] f d F or F n...() so tht () cn e written s Clculus Get Discount Coupons for your Coching institute nd FREE Study Mteril t

8 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus In words, the theorem tells us tht Emple 7. Find Solution : f ( ) d (Vlue of ntiderivtive t the upper limit ) (Vlue of the sme ntiderivtive t the lower limit ) d d Emple 7. Evlute the following () cos d Solution : We know tht () cosd sin + c cosd [ sin] () e d sin sin e e d, e d e Definite Integrls e Theorem : If f nd g re continuous functions defined in [, ] nd c is constnt then, (i) ( ) ( ) c f d c f d o (ii) [ ( ) + ( )] ( ) + ( ) f g d f d g d (iii) [ ( ) ( )] ( ) ( ) f g d f d g d MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

9 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Emple 7.5 Evlute ( + ) Solution : ( ) 5 7 d d d 5 d + 7 d d 5 d + 7 d Clculus CHECK YOUR PROGRESS S 7. 5 MATHEMATICS ( ). Find ( + ) d s the limit of sum.. Find. Evlute () (c) [ ] sin d () ( sin + cos ) d e d s the limit of sum. (d) ( + + ) d d 7. EVALUATION ATION OF DEFINITE INTEGRAL BY SUBSTITUTION UTI ww. The principl step in the evlution of definite integrl is to find the relted indefinite integrl. In the preceding lesson we hve discussed severl methods for finding the indefinite integrl. One of the importnt methods for finding indefinite integrls is the method of sustitution. When we use sustitution method for evlution the definite integrls, like d +, sin d, + cos Get Discount Coupons for your Coching institute nd FREE Study Mteril t

10 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus Definite Integrls the steps could e s follows : (i) Mke pproprite sustitution to reduce the given integrl to known form to integrte. Write the integrl in terms of the new vrile. (ii) Integrte the new integrnd with respect to the new vrile. (iii) Chnge the limits ccordingly nd find the difference of the vlues t the upper nd lower limits. Note : If we don't chnge the limit with respect to the new vrile then fter integrting resustitute for the new vrile nd write the nswer in originl vrile. Find the vlues of the nswer thus otined t the given limits of the integrl. Emple 7.6 Evlute + d Solution : Let + t d dt or d dt When, t 5 nd, t. Therefore, e, 5 nd re the limits when t is the vrile. Thus d dt t + 5 [ logt ] 5 log log5 [ ] log Emple 7.7 Evlute the following : () sin wwsin d yc ww+ + cos sin θ + cos θ Solution : () Let cos t then sin d dt () sinθ dθ (c) ww. w.pick d 5 + cos When, t nd, t. As vries from to, t vries from to. sin + cos + t d dt [ tn t] tn tn MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

11 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Clculus () sinθ sinθ I d θ d θ θ + θ θ + θ θ θ ( ) sin cos sin cos sin cos sin θ d θ sinθcos θ sinθdθ sin θ sin θ Let sin θ t Then sin θ cos θ dθ dt i.e. sinθ dθ dt ( ) When θ,t nd θ, t. As θ vries from to c,, the new vrile t vries from to. I t ( t ) dt dt t t + I dt t t + + I.p t + ww.pic t. tn ( ) tn tn toc MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t dt

12 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus tn (c) We know tht cos + tn Let Then d 5 + cos tn t 5 + d tn sec 9 + tn ( + tn ) Definite Integrls d () sec d dt when, t, when d 5 + cos 9 + t dt [From ()] M t tn tn, t 7. SOME PROPERTIES OF DEFINITE INTEGRALS The definite integrl of f () etween the limits nd hs lredy een defined s f ( ) d F( ) F( ), Where [ ( )] ( ) d F f, d www. ww.pickmycoching.com ing.c om where nd re the lower nd upper limits of integrtion respectively. Now we stte elow some importnt nd useful properties of such definite integrls. (i) ( ) ( ) f d f t dt (ii) f ( ) d f ( ) d (iii) ( ) ( ) + ( ) ycdt c f d f d f d, where <c<. c MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

13 w( Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls (iv) ( ) ( + ) f d f d (v) ( ) ( ) + ( ) f d f d f d (vi) ( ) ( ) (vii) (viii) f d f d ( ) ( ), if f f f ( ) d f d, if f f ( ) ( ) ( ) ( ), if f isnoddfunctionof of f ( ) d f ( ) d, if f ( ) isnevenfunctionof nctionof Mny of the definite integrls my e evluted esily with the help of the ove stted properties, which could hve een very difficult otherwise. The use of these properties in evluting definite integrls will e illustrted in the following emples. Emple 7.8 Show tht () Solution : () Let log tn d () (ww ww f ( ) d Using the property ( ) ( ) d + sin I log g t tn d...(i) (ww ww.p w.p f d f d,weget I log tn d km( Clculus log ( cot ) d MATHEMATICS 5 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

14 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus ( ) log tn d Definite Integrls 6 logtnd I [Using (i)] I i.e. I or () Let d + sin I d + sin log g tn d I d ( ) ( ) + sin ( ) Adding (i) nd (ii) d + sin f d f d + I d d + sin + sin sin or I d sin w.pic com ( ) sec tnsec d [ tn ] [( tn sec ) ( tn sec) ] [ ( ) ( ) ] sec I Cod (i) (ii) MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

15 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Emple 7.9 Evlute () sin d sin + cos () sin cos d + sincos Clculus Solution : () Let I sin d sin + cos (i) Also I Adding (i) nd (ii), we get i.e. () Let I sin d sin cos + I sin cos d + sincos.d (Using the property ( ) d ( ) cos d cos + sin sin + sin + [ ] cos d cos I sin d sin + cos MATHEMATICS 7 Get Discount Coupons for your Coching institute nd FREE Study Mteril t f d f d ). hing. + kmycoching.com (ii) (i)

16 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus 8 Then I sin cos d + sin cos Definite Integrls ( ) ( ) Adding (i) nd (ii), we get cos sin d + cossin sin cos cos sin I + + sincos d + sincos I Emple 7. Evlute () Solution : () Here f ( ) f ( ) sin cos + cos sin n d + sincos e + e ( + e + f ( ) f is n odd function of. ( ) ) () + d e + d d Co od () +,if +,if < f d f d + d ing.c co om + d + d + + d, using property (iii) (ii) MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

17 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls ( ) ( ) d + + d + + Emple 7. Evlute ( ) Solution : Let I ( ) Also log sin d log sin d I log sin d, Adding (i) nd (ii), we get log ( cos ) d [ ( ) +ckmcos log ( cos )] I log sin + log cos d ( sincos ) log ( sincos ww sin logwg ws w d d ( ) ( ) log sin d log d log ( sin ) d log [Using property (iv)] low ww.p ching (i) (ii) (iii) Clculus MATHEMATICS 9 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

18 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus Agin, let I ( ) log sin d Put t d dt When, t nd,t I ( ) log sint dt Definite Integrls. log ( sint ) dt, [using property (vi)]. log ( sin ) dt [using property (i)] I I, [from (i)]...(iv) Putting this vlue in (iii), we get I I log I log Hence, ( ) log sin d log CHECK YOUR PROGRESS 7. Evlute the following integrls : e d. 5 + d 5. 5 log cosd 8. cos d + sin + cos d. 5 + sin ww.pick d 6. e d d 5 + sin d cos + sin sinlogtnd MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

19 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls 7. APPLICATIONS OF INTEGRATION Suppose tht f nd g re two continuous functions on n intervl [, ] such tht f ( ) g( ) for [, ] tht is, the curve y f () does not cross under the curve y g () over [, ]. Now the question is how to find the re of the region ounded ove y y f (), elow y y g (), nd on the sides y nd. Agin wht hppens when the upper curve y f () intersects the lower curve y g () t either the left hnd oundry, the right hnd oundry or oth? Clculus 7.. Are Bounded y the Curve, -is nd the Ordintes Let AB e the curve y f () nd CA, DB the two ordintes t nd respectively. Suppose y f () is n incresing function of in the intervl. Let P (, y) e ny point on the curve nd Q ( + δ, y +δ y) neighouring point on it. Drw their ordintes PM nd QN. Here we oserve tht s chnges the re (ACMP) lso chnges. Let AAre (ACMP) Then the re (ACNQ) A +δ A. The re (PMNQ)Are (ACNQ) Are (ACMP) A + δa A δ A. Fig.7.6 Complete the rectngle PRQS. Then the re (PMNQ) lies etween the res of rectngles PMNR nd SMNQ, tht is δ A lies etween y δ nd ( y + δy) δ δa lies etween y +δy) δ w y nd ( In the limiting cse when wq Q P, δ nd δy. A. ww.pi w.pickm ickmycoching.com ing.c com lim δ δa δ lies etween y nd lim ( y +δy) δy da y d Integrting oth sides with respect to, from to, we hve MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

20 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus da yd d A d [ ] Hence Are (ACDB) ( ) f d (Are when ) (Are when ) Are (ACDB) Are (ACDB). Definite Integrls The re ounded y the curve y f (), the -is nd the ordintes, is ( ) f d or yd where y f () is continuous single vlued function nd y does not chnge sign in the intervl. Emple 7. Find the re ounded y the curve y, -is nd the lines,. Solution : The given curve is y Required re ounded y the curve, -is nd the ordintes, (s shown in Fig. 7.7) is d squre units Emple 7. Find the re ounded y the curve quw w.pfi ww wwt ckm km y e, -is nd the ordintes nd >. Solution : The given curve is y e. ww.pi w.p ching Required re ounded y the curve, -is nd the ordintes, is Fig. 7.7 e d e ( ) e squre units MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

21 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Emple 7. Find the re ounded y the curve y,, c. Solution : The given curve is y ccos c Required re yd ccos d c c sin c c sin sin c c sin squre units c ccos, -is nd the ordintes c Emple 7.5 Find the re enclosed y the circle + y, nd -is in the first qudrnt. Solution : The given curve is + y, which is circle whose centre nd rdius re (, ) nd respectively. Therefore, we hve to find the re enclosed y the circle + y, the -is nd the ordintes nd. Required re yd ww d, is positive in ( y is positive in the first qudrnt) + sin + sin sin. sin,sin squre units ww. w.pic www Fig. 7.8 Clculus MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

22 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus Definite Integrls Emple 7.6 Find the re ounded y the -is, ordintes nd the following curves : (i) y c,,, > > (ii) y loge,,, > > Solution : (i) Here we hve to find the re ounded y the -is, the ordintes, nd the curve y c or Are yd ( > given) (ii) Here Are c d c [ log] ( ) c log log c log y log e loge d, ( > > ) [ M] ] log e d log e e log log d [ ] log log e e log log + e km e ( ) ( ) log log e loge log e e e e y c ( log e ) e CHECK YOUR PROGRESS 7.. Find the re ounded y the curve y, -is nd the lines,.. Find the re ounded y the curve y, -is nd the lines nd. MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

23 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls. Find the re ounded y the curve y e, -is nd the ordintes,, >.. Find the re ounded y the -is, the curve nd, c. y c sin nd the ordintes c 7... Are Bounded y the Curve f (y) etween y-is nd the Lines y c, y d Clculus Let AB e the curve f (y) nd let CA, DB e the scisse t y c, y d respectively. Let P (, y) e ny point on the curve nd let Q ( + δ, y +δ y) e neighouring point on it. Drw PM nd QN perpendiculrs on y-is from P nd Q respectively. As y chnges, the re (ACMP) lso chnges nd hence clerly function of y. Let A denote the re (ACMP), then the re (ACNQ) will e A +δ A. The re (PMNQ) Are (ACNQ) Are (ACMP) A + δ A A δ A. Complete the rectngle PRQS. Then the re (PMNQ) lies etween the re (PMNS) nd the re (RMNQ), tht is, δ A lies etween y + δ δy δ nd ( ) δa lies etween nd + δ δy In the limiting position when Q P, δ nd δy. δa lim lies etween nd lim ( +δ ) δ y δ y δ da dy ww Integrting oth sides with respect to y, etween the limits c to d, we get d d dy dy wda wwda dy c dyww c [ A ] d c (Are when y d) (Are when y c) Are (ACDB) Are (ACDB) Hence re (ACDB) dy f ( y) dy d c d c Fig ww ching ing.c ng om wwd MATHEMATICS 5 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

24 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus Definite Integrls The re ounded y the curve f ( y ), the y-is nd the lines y c nd y d is d dy or f ( y) dy c d c where f ( y ) is continuous single vlued function nd does not chnge sign in the intervl c y d. Emple 7.7 Find the re ounded y the curve y, y-is nd the lines y, y. Solution : The given curve is y. Required re ounded y the curve, y-is nd the lines y, y is dy y dy y 9 9 squre units Emple 7.8 Find the re ounded y the curve y, y is nd the lines y, y. Solution : The eqution of the curve is y Required re ounded y the curve, y-is nd the lines y, y y y dy ww.p Fig squre units Emple 7.9 Find the re enclosed y the circle + y nd y-is in the first qudrnt. 6 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

25 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Solution : The given curve is + y, which is circle whose centre is (, ) nd rdius. Therefore, we hve to find the re enclosed y the circle + y, the y-is nd the scisse y, y. Clculus Required re dy y dy (ecuse is positive in first qudrnt) y y y sin + + sin sin squre units,sin sin,sin Note : The re is sme s in Emple 7., the reson is the given curve is symmetricl out oth the es. In such prolems if we hve een sked to find the re of the curve, without ny restriction we cn do y either method. MATHEMATICS 7 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Emple 7. Find the whole re ounded y the circle + y. Solution : The eqution of the curve is + y. The circle is symmetricl out oth the es, so the whole re of the circle is four times the re os the circle in the first qudrnt, tht is, Are of circle re of OAB (From Emple 7.5 nd 7.9) squre units Emple 7. Find the whole re of the ellipse y + Solution : The eqution of the ellipse is y + Fig w.pick com Fig. 7.

26 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus The ellipse is symmetricl out oth the es nd so the whole re of the ellipse is four times the re in the first qudrnt, tht is, Whole re of the ellipse re (OAB) In the first qudrnt, y or y Now for the re (OAB), vries from to Are (OAB) yd d Hence the whole re of the ellipse + sin + sin sin. squre units 7.. Are etween two Curves Definite Integrls Suppose tht f () nd g () re two continuous nd non-negtive functions on n intervl [, ] such tht f ( ) g( ) for ll [, ] tht is, the curve y f () does not cross under the curve y g () for [,]. We wnt to find the re ounded ove y y f (), elow y y g (), nd on the sides y nd. ing.c.com Let A [Are under y f ()] [Are under y g ()]...() Now using the definition for the re ounded co Fig. 7. y the curve y f ( ), -is nd the ordintes nd, we hve Are under Fig.7. 8 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

27 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls ( ) ( ) y f f d Similrly,Are under ( ) ( ) Using equtions () nd () in (), we get...() y g g d...() ( ) ( ) A f d g d [ ( ) ( )] f g d...() Wht hppens when the function g hs negtive vlues lso? This formul cn e etended y trnslting the curves f () nd g () upwrds until oth re ove the -is. To do this let-m e the minimum vlue of g () on [, ] (see Fig. 7.5). Since g ( ) m ( ) g + m Fig. 7.5 Fig. 7.6 Now, the functions g ( ) + m f ( ) + mre non-negtive on [, ] (see Fig. 7.6). It is intuitively cler tht the re wwnd of region is unchnged y trnsltion, so the re A etween f nd g is the sme s the re etween g ( ) + m nd f ( ) + m. Thus, ycoching.co g.com [ ] A [ re under y [ f ( ) + m] ] [re under y g ( ) + m ]...(5) Now using the definitions for the re ounded y the curve y f (), -is nd the ordintes nd, we hve Clculus Are under ( ) + [ ( ) + ] y f m f m d...(6) MATHEMATICS 9 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

28 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus nd Are under The equtions (6), (7) nd (5) give ( ) [ ( ) ] Definite Integrls y g + m g + m d (7) [ ( ) ] [ ( ) ] A f + m d g + m d [ ( ) ( )] f g d which is sme s () Thus, If f () nd g () re continuous functions on the intervl [, ], nd f () g (), [, ], then the re of the region ounded ove y y f (), elow y y g (), on the left y nd on the right y is [ ( ) ( )] f g d Emple 7. Find the re of the region ounded ove y y + 6, ounded elow y y, nd ounded on the sides y the lines nd. Solution : y + 6 is the eqution of the stright line nd y is the eqution of the prol which is symmetric out the y-is nd origin the verte. Also the region is ounded y the lines nd. ching.com Thus, ( ) Fig. 7.7 A + 6 d d + 6 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

29 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls squre units If the curves intersect then the sides of the region where the upper nd lower curves intersect reduces to point, rther thn verticl line segment. Emple 7. Find the re of the region enclosed etween the curves y nd y + 6. Clculus Solution : We know tht y is the eqution of the prol which is symmetric out the y-is nd verte is origin nd y + 6 is the eqution of the stright line which mkes n ngle 5 with the -is nd hving the intercepts of 6 nd 6 with the nd y es respectively. (See Fig. 7.8). Fig. 7.8 A sketch of the region shows tht the lower oundry is y nd the upper oundry is y +6. These two curves intersect t two points, sy A nd B. Solving these two equtions we get ( ) ( + ), When, y 9 nd when, y ( The required re ( ( + ) + 6 d squre units 6 Emple 7. Find the re of the region enclosed etween the curves y nd y. Solution : We know tht y is the eqution of the prol which is symmetric out the MATHEMATICS ww.p w.pic Get Discount Coupons for your Coching institute nd FREE Study Mteril t

30 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus Definite Integrls y-is nd verte is origin. y is the eqution of the stright line pssing through the origin nd mking n ngle of 5 with the -is (see Fig. 7.9). A sketch of the region shows tht the lower oundry is y nd the upper oundry is the line y. These two curves intersect t two points O nd A. Solving these two equtions, we get ( ), Here f ( ), g ( ), nd Therefore, the required re ( ) d squre units 6 Emple 7.5 Find the re ounded y the curves y nd y. Solution : We know tht y the eqution of the prol which is symmetric out the -is nd origin is the verte. e. y is the eqution of the stright line pssing through origin nd mking n ngle of 5 with the -is (see Fig. 7.). A sketch of the region shows tht the lower oundry is y nd the upper oundry is y. These two curves intersect t two points O nd A. Solving these two equtions, we get y y y ( y ) y, When y, nd when y,. ww. w. om Here ( ) ( ) ( ) Therefore, the required re is f, g,, ( ) d Fig. 7.9 Fig. 7. MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

31 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls 8 8 squre units Emple 7.6 Find the re common to two prols Solution : We know tht y nd re symmetric out the -is nd y-is respectively. y nd y. y re the equtions of the prols, which Also oth the prols hve their vertices t the origin (see Fig. 7.9). Clculus A sketch of the region shows tht the lower oundry is y nd the upper per oundry is y. These two curves intersect t two points O nd A. Solving these e two equtions, we hve 6 ( ) 6, Hence the two prols intersect t point (, ) nd (, ). f,g, nd Therefore, required re Here ( ) ( ) w.p d. 6 6 squre units Fig w.pic ching.co com MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

32 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus CHECK YOUR PROGRESS 7.. Find the re of the circle + y 9 Definite Integrls. Find the re of the ellipse. Find the re of the ellipse y + 9 y Find the re ounded y the curves y ndy y 5. Find the re ounded y the curves y nd y. om 6. Find the re enclosed y the curves y nd y + LET US SUM UP If f is continuous in [, ] nd F is n nti derivtive of f in [, ], then ( ) ( ) F ( ) f d F F If f nd g re continuous in [, ] nd c is constnt, then (i) ( ) d c f( ) LET US SUM UP c f d c f d (ii) [ f ( ) + ( )] ( ) + ( ) f g d f d g d (iii) i) [ f ( ) ( )] ( ) ( ) f g d f d g d ww.pic ckmy kmycoching.com om i) ww The re ounded y the curve y f (), the -is nd the ordintes is ( ), f d or where y f ( ) is continuous single vlued function nd y does not chnge sign in the intervl yd MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

33 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls If f () nd g () re continuous functions on the intervl [, ] nd f ( ) g( ), for ll [, ], then the re of the region ounded ove y y f (), elow y y g (), on the left y nd on the right y is [ ( ) ( )] f g d Clculus TERMINAL EXERCISE Evlute the following integrls ( to 5) s the limit of sum... d. d. cosd 5. ( + ) d Evlute the following integrls (6 to 5) SUPPORTIVE WEB SITES d 7. cos d. sind 8. sin d. sind cotd ww.pi w.pic ch d. d. dθ 5 + cos θ. tn d 5. sin d 6. + d 7. 5 sin θ cos θdθ MATHEMATICS 5 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

34 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Clculus 8.. logsind 9. log ( + ) sin d sin + cos sin cos d. d + cos. log ( + ). ( + ) d tn d. 8 5 Definite Integrls sin cosd 6 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t

35 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls ANSWERS Clculus CHECK YOUR PROGRESS e e. () () (c) (d) 6 CHECK YOUR PROGRESS 7.. e. tn log6 + tn log log CHECK YOUR PROGRESS sq. units. 7 sq. units e. sq. units c cos c CHECK YOUR PROGRESS sq. units ww. 6 sq. units. sq. units 6. sq. units 5. 6 sq. units 6. 9 sq. units TERMINAL EXERCISE MyC w.... cos cos. sin sin MATHEMATICS 7 Get Discount Coupons for your Coching institute nd FREE Study Mteril t

36 Get Discount Coupons for your Coching institute nd FREE Study Mteril t Definite Integrls Clculus log log 5.. log ( ) 6 8. log 9. log.. log log ( + ) com. 78 mlog com 8 MATHEMATICS Get Discount Coupons for your Coching institute nd FREE Study Mteril t