Mathematics. Area under Curve.

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Mthemtics Are under Curve www.testprepkrt.com

Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding the Ares using Integrtion. 6. Symmetricl Are. 7. Are etween Two curves. 8. Volumes nd Surfces of Solids of Revolution. 1

1. Introduction. We know the methods of evluting definite integrls. These integrls re used in evluting certin types of ounded regions. For evlution of ounded regions defined y given functions, we shll lso require to drw rough sketch of the given function. The process of drwing rough sketch of given function is clled curve sketching.. Procedure of Curve Sketching. (1) Symmetry: (i) Symmetry out -is: If ll powers of y in eqution of the given curve re even, then it is symmetric out -is i.e., the shpe of the curve ove -is is ectly identicl to its shpe elow -is. For emple, y 4 is symmetric out -is. (ii) Symmetry out y-is: If ll power of in the eqution of the given curve re even, then it is symmetric out y-is For emple, 4y is symmetric out y-is. (iii) Symmetry in opposite qudrnts or symmetry out origin: If y putting for nd y for y, the eqution of curve remins sme, then it is symmetric in opposite qudrnts. For emple, y nd y re symmetric in opposite qudrnts. (iv) Symmetry out the line nd y then it is symmetric out the line 0 45 with the positive direction of -is. y : If the eqution of given curve remins unltered y interchnging y which psses through the origin nd mkes n ngle of

() Origin: If the eqution of curve contins no constnt terms then it psses through the origin. Find whether the curve psses through the origin or not. For emples, y 4 0 psses through origin. (3) Points of intersection with the es: If we get rel vlues of on putting y 0 in the eqution of the curve, then rel vlues of nd y 0 give those points where the curve cuts the -is. Similrly y putting 0, we cn get the points of intersection of the curve nd y-is. y For emple, the curve 1 intersect the es t points (, 0) nd ( 0, ). dy (4) Specil points: Find the points t which 0, t these points the tngent to the curve is prllel to d d -is. Find the points t which 0. At these points the tngent to the curve is prllel to y-is. dy (5) Region: Write the given eqution s y f(), nd find minimum nd mimum vlues of which determine the region of the curve. For emple for the curve y ( ) y Now y is rel, if 0, So its region lies etween the lines = 0 nd = (6) Regions where the curve does not eist: Determine the regions in which the curve does not eists. For this, find the vlue of y in terms of from the eqution of the curve nd find the vlue of for which y is imginry. Similrly find the vlue of in terms of y nd determine the vlues of y for which is imginry. The curve does not eist for these vlues of nd y. For emple, the vlues of y otined from y 4 re imginry for negtive vlue of, so the curve does not eist on the left side of y-is. Similrly the curve y ( ) does not eist for s the vlues of y re imginry for. 3

3. Sketching of Some Common Curves. (1) Stright line: The generl eqution of stright line is y c 0. To drw stright line, find the points where it meets with the coordinte es y putting y = 0 nd = 0 respectively in its eqution. By joining these two points, we get the sketch of the line. () Region represented y liner inequlity: To find the region represented y liner inequlities y c nd y c, we proceed s follows. (i) Convert the inequlity into equlity to otin liner eqution in, y. (ii) Drw the stright line represented y it. (iii) The stright line otined in (ii) divides the y-plne in two prts. To determine the region represented y the inequlity choose some convenient points, e.g. origin or some point on the coordinte es. If the coordintes of point stisfy the inequlity, then region contining the point is the required region, otherwise the region not contining the point is the required region. 1 (3) Circle: The eqution of circle hving center t (0,0) nd rdius r is given y y r. The eqution of circle hving center t (h, k) nd rdius r is given y ( h) ( y k) r. The generl eqution of circle is 1 1 1 y g fy c 0. This represents the circle whose center is t (-g,-f) nd rdius equl to g f c. The figure of the circle y () is given. Here center is (0,0) nd rdius is. 4

(4) Prol: There re four stndrd forms of prol with verte t origin nd the is long either of coordinte is. (i) y 4 : For this prol () Verte: (0,0) () Focus: (, 0) (c) Directri: 0 (d) Ltus rectum: 4 (e) Ais y 0 (f) Symmetry: It is symmetric out -is. Directri Z A S (,0) (ii) 4y : For this prol () Verte: (0,0) () Focus: ( 0, ) (0, ) (c) Directri: y 0 (d) Ltus rectum: 4 (e) Ais = 0 (f) Symmetry: It is symmetric out y-is ' Z A Directri ' (5) Ellipse: The stndrd eqution of the ellipse hving its center t the origin y nd mjor nd minor es long the coordinte es is 1 Here. The figure of the ellipse is given. B (0,) (-, 0) (, 0) A A B (0, ) 5

4. Are of Bounded Regions. (1) The re ounded y Crtesin curve y = f(), -is nd ordintes = nd = is given y Are y d f( ) d y = f() O y d = = () If the curve y = f() lies elow -is, then the re ounded y the curve y f(), the -is nd the ordintes = nd = is negtive. So, re is given y y d (3) The re ounded y Crtesin curve =f(y), y-is nd sciss y = c nd y = d is given y d Are = c c dy f( y) dy d y = d y = c dy = f(y) O (4) If the eqution of curve is in prmetric form, sy = f(t), y = g(t) then the re t t y d g( t) f' ( t) dt where t 1 nd t re the vlues of t respectively corresponding to the vlues of nd of. 1 6

5. Sign convention for finding the Ares using Integrtion. While pplying the discussed sign convention, we will discuss the three cses. Cse I: In the epression f ( ) d if nd f ( ) 0 for ll, then this integrtion will give the re enclosed etween the curve f(), -is nd the line = nd = which is positive. No need of ny modifiction. Cse II: If in the epression f ( ) d if > nd f ( ) 0 for ll, then this integrtion will clculte to e negtive. But the numericl or the solute vlue is to e tken to men the re enclosed etween the curve y = f(), -is nd the lines = nd =. Cse III: If in the epression f ( ) d where ut f() chnges its sign numers of times in the intervl, then we must divide the region [, ] in such wy tht we clerly get the points lying etween [, ] where f() chnges its sign. For the region where f()>0 we just integrte to get the re in tht region nd then dd the solute vlue of the integrtion clculted in the region where f()<0 to get the desired re etween the curve y = f(), -is nd the line = nd =. Hence, if f() is s in ove figure, the re enclosed y y = f(), -is nd the lines = nd = is given y c d e A f( ) d f( ) d f( ) d f( ) d f( ) d c d e f f c d e f 7

6. Symmetricl Are. If the curve is symmetricl out coordinte is (or line or origin), then we find the re of one symmetricl portion nd multiply it y the numer of symmetricl portions to get the required re. 7. Are etween Two curves. (1) When oth curves intersect t two points nd their common re lies etween these points: If the curves y f ) nd y f ), where f ) f ( ) intersect in two points 1 1 ( ( 1( ( 1 A( = ) nd B( = ), then common re etween the curves is y y ) d y = f 1 () B A y = f () [ 1 = f ( ) f ( )] d O = d = () When two curves intersect t point nd the re etween them is ounded y -is: Are ounded y the curves y f1 ( ), y f ( ) nd is is = 1 ) d f f ( ) d ( y 1 = f 1 () P(α,β) y = f () Where P(, ) is the point of intersection of the two curves. O (3) Positive nd negtive re : Are is lwys tken s positive. If some prt of the re lies ove the -is nd some prt lies elow -is, then the re of two prts should e clculted seprtely nd then dd their numericl vlues to get the desired re. 8

Importnt Tips The re of the region ounded y y 4 nd 16 4 y is squre units. 3 The re of the region ounded y y 4 nd y m is squre units 3 The re of the region ounded y 8 3m y 4 nd its ltus rectum is 8 3 squre units The re of the region ounded y one rch of sin () or cos () nd -is is sq. units Are of the ellipse 1 is sq. units. y Are of region ounded y the curve y = sin, -is nd the line =0 nd = is 4 unit. 8. Volumes nd Surfces of Solids of Revolution. If plne curve is revolved out some is in the plne of the curve, then the ody so generted is known s solid of revolution. The surfce generted y the perimeter of the curve is known s surfce of revolution nd the volume generted y the re is clled volume of revolution. For emple, right ngled tringle when revolved out one of its sides (forming the right ngle) genertes right circulr cones. Volumes of solids of revolution: (i) The volume of the solid generted y the revolution, out the -is, of the re ounded y the curve y = f(), the ordintes t =, = nd the -is is equl to y d. (,y) A R P Q (+, y+y) S = = O K N M L (ii) The revolution of the re lying etween the curve = f(y), the y-is nd the lines given y (interchnging nd y in the ove formule) dy. y nd y is (iii) If the eqution of the generting curve e given y f 1 ( t ) nd y f ( t) nd it is revolved out - is, then the formul corresponding to y d ecomes { f ( t)} d { f1 ( t)}, where f 1 nd f re t the vlues of t corresponding to = nd = t 1 9

(iv) If the curve is given y n eqution in polr co-ordintes, sy r f( ), nd the curve revolves out the initil line, the volume generted y d d y. d, where nd re the vlues of corresponding to = nd = d Now r cos, y r sin. Hence the volume d r sin ( r cos ) d d (v) If the generting curve revolves out ny line AB (which is different from either of the es), then the volume of revolution is ( PN ) d ( ON ), P Q D C A O N M B Note: The volume of the solid generted y revolving the re ounded y the curve r f( ) nd the rdii vectors 3 nd out the initil line is r sin d. 3 The volume in the cse when the ove re is revolved out the line is 3 r 3 cos d. () Are of surfces of revolution: (i) The curved surfce of the solid generted y the revolution, out the - is, of the re ounded y the curve y f(), the ordintes t =, = nd the -is is equl to y ds. C S Q A B P R = = O D N M E (ii) If the rc of the curve y f() revolves out y-is, then the re of the surfce of revolution dy (etween proper limits) = ds, where ds 1 d d (iii) If the eqution of the curve is given in the prmetric form f 1( t ) nd y f ( t), nd the curve tt t revolves out -is, then we get the re of the surfce of revolution yds tt t t f 1 t1 ( t) ds 10

t d dy f ( t) dt, where t 1 nd t re the vlues of the prmeter t corresponding to = t1 dt dt nd =. (iv) If the eqution of the curve is given in polr form then the re of the surfce of revolution out - ds dr y ( r sin ). d r r d d sin. etween proper limits. d is ds 11