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Applications of the Integrals C Fastracl« Revision ) > IF F(x) 20 Is a continuous Function which is defined in the > Area of the region bounded by the curve x=f(y), Y-axis, interval (a, 5), then the area of the region bounded by the y=cand y=d Is given by d d curve y=/(x), Xaxis, x=aand x =b Is given by Area =f! xdy =f F(y)ay Area =| ® Fix)dx =| Pde > IF the curve x =f(y), lies on left of Y-axis, then area of the = i region bounded by curve x= f(y), Y-axis, y=cand y=d is Here, curve lies above X-axis. given by Area= d d > IF F(x) <0 is a continuous function which is defined in the j: xdy (f° fy) 4y| (take numerical value) Interval (a, 6), then area of thereglon bounded bythecurve iF for x e[a,c], f(x)20 and for xe[c,b],f(x) 0, where y =F(x), X-axis, x =a and x =bis given by a is equal to: a. rq. units b. 1sq. unit c. 2 If the area of the region bounded by the lines y=mx,x=1, x =2and X-axis, is 6 sq. units, then mis equal to: a3 b.1 CZ d 4 The area in the positive quadrant enclosed by the circle x? +y? =4, the line x = y/3 and X-axis, is: (CBSE 2022 Terni-2) Re Sq. units 4 d. 2 sq. units ig. units iq. units, c. 559 units d. sq. units y Assertion & Reason Type Questions Directions (Q. Nos. 21-25): /n the following questions, each question contains Assertion (A) and Reason (R). Each question has 4 choices (a), (b), (¢) and (d) out of which only one is carrect. The choices are: Q21. Q 22. Q 23. Q 24. Q 26. Q20. The area enclosed by y =3x -5, y =0, x =3 and x 5, is: a. 12 sq. units b. 13 sq. units C. 5sq units c. 14 sq. units Answers 1. (ob) 2. (b) 3. (a) 4. (a) 5. (a) 6. () 11. (b) 12. (b) 12. (d) 14, (b) 18. (a) 16. (b) 21. (a) 22. (b) 23. (a) 24. (c) 26. (D) a. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A) Both Assertion (A) and Reason (R) are true but Reason (R) Is not the correct explanation of Assertion (A) c. Assertion (A) is true and Reason (R) Is false d. Assertion (A) is false and Reason (R) Is true Assertion (A): The area of the region bounded by the curve y? = 4x and the line x =3is 83 sq. units. Reason (R): If f(x) 20 is a continuous function which is defined in the interval [a, 6], then the area of the region bounded by the curve y = f(x), X-axis, x =a and x =b is given by Area =|. f(x) dx = fy de o Assertion (A): The area bounded by the pa rabola 56a y? =4axand the lines x =aand x =4ais sq. units. , x? y? . Reason (R): Area of an ellipse ='SF lis nab a ob sq. units. Assertion (A): The area enclosed by the curve Ixl+ly|=2is 8 units. Reason (R): |x|+|y|=2 represents a square of side length V8 units. Assertion (A): The area bounded by the curve y =2cosx and the X-axis from x =Oto x =2zis 8 sq. units. Reason (R): The area bounded by the curve y =sinx between x =Qand x =2n is 2 sq. units. Assertion (A): The area of the region in the first quadrant, bounded by the parabola y = 9x? and the lines x =0,y =1andy =4is 14/9 sq. units. Reason (R): If for x e[a,c], f(x)20 and for x e[c,b], f(x) <0, where a 4y? =4 y? = yodl From eq. (2), x = V3, ) *. Points of intersection of pizza and edge of knife are (8.1) (8. -1) So, option (b) Is correct. 2. So, option (a) Is correct. ix 2 z 3. Required area = [) BX ls 4x? dx -3[f) ERP tor Ql, “9/34 [psr'0- ie Se 3) ale fh 2 28 Beg nits 5 . So. aption (b) is correct: 4. We have, xt4y2=4 = (x-0)? +(y-0)? = (2)? Z Radius =2 Area of ah slice of pizza = + (2)? =1sq. units So. aption (a) Is correct. 5. Area of whole pizza = x (2)? = 4xsq. units So, option (d) is correct Case Study 3 In a classroom, teacher explains the properties of a particular curve by saying that this particular curve has beautiful ups and downs. It starts at 1 and heads down until 2 radian and then heads up again and closely telnte to sine function and both follow each other, exattly = radians apart as shown in figure. [ = sink ~2n/ = 7 wit SY, i > | / / i / Y cA A Sy, A 2 Based on the above information, solve the following questions: Q 1. Write the name of curve, about which teacher explained in the classroom. Q 2. Find the area of curve explained in the above passage from 0 to 5. Q 3. Find the area of que discussed in the above assage from = to >S Passag 2 2 Or Find the area of curve discussed in the above Passage from = to2n. ° Solutions ° 1. Here, teacher explained about cosine curve. 2 2. Required area= ipa cos x dx =(sin xf? =sin5 -sinO=1-0=1 sq. unit De ee Bw2 3. Required area = [fos cosx ck | =I (sin x)2771 = sin3% sin= 2 =|-1-1l=|-2|=25q. units (Since, area can't be negative) Or 2n Required area = fous cos x dx =(sin x]§)> =sin2r -sin= = Q-(-1) =1sq. unit Case Study 4 Graphs of two functions f(x) =sin x and g(x) =cosx is given below: g(x) = cos x f(x)=ssinx Ns a Based on the above information, solve the following questions: Q1. In [0, 1], the curves f(x) =sin x and g(x) =casx intersect at x = Ea t i as b. 5 cin dr Q2. The value of fr" sin x dx is: aie b+) e2- d242 Q 3. The value of (aa cos x dx is: a. 1+ b.1- c2- d 2+ Bb ch dae Q 4. The value of | sin x dx is: a0 b.1 G2 d.-2 Q6. The value off" sin x dx is: aO b.1 G2 d.=2 o——_—————_ Solutions 1. For point of Intersection, we have sin x =cos x sinx =] cosXx > > tanxel=> xem “4 So, option (c) Is correct. 2." sin x dx =(-cos x}f* =-cos=+cos0 =1- I 4 vz So. option (a) Is correct. 3. (me cos x dx =(sin x}¥2 1 esine ogy Sule 24 So. option (b) is correct. 4 i sin x dx =(- cos x]g =[-cosx+cosO] =(1+1)= So, option (c) is correct. 7] 5. fe sinx dx =(-cosx] 7 = [-c053+-cos0 O+1=1 So, option (b) Is correct. Case Study 5 Consider the following, equations of curves x? = y and y=x. Based on the above information, solve the following questions: Q1. Find the point(s) of intersection of both the curves. Q 2. Draw the graph of area bounded by the curves. Q 3. Find the value of the integral { ° x dx. Or Find the value of the integral in x? dx. ° Solutions © 1. We have, xtey (I) and X=y si(2) From eqs. (1) and (2), xrex => xtoxsd > x(x-1)=0 = x=0,1 From eq. (2), y=0.1 Thus, required points of Intersection are (0, 0), (1, 1). 2a iva “a Very Short Answer type Questions Q1. Find the area of the region bounded by the curve y =x? and the line y =4 (NCERT EXERCISE) Q2. Find the area of the region bounded by the parabola y” = 8x and the line x =2 (NCERY EXEMPLAR; CBSE 2020) Qa. Using integration, find the area of the region bounded by lines x -y +1=0, x =-2, x =3 and X-axis. (CBSE 2022 Term-2) Q4. Find the area bounded by the curve x =2y +3, Y-axis and the lines y = Land y =-1. (NCERT EXEMPLAR) Q5. Find the area of the region enclosed by the curves y’=x,x ae y =Oand x = Lusing integration. 2 (CBSE 2022 Term-2) 06. Find the area of the region bounded by y? =9x, x =2, x =4and the X-axis in the first quadrant. (NCERT EXERCISE) Q7. Find the area bounded by y = x’, X-axis and the lines x =-1and x =1. Q8. Find the area between O and x of the curve y =sinx. (NCERT EXEMPLAR) Q 9. Find the area of the region bounded by the curve y =cosx between x =Oand x =n. (NCERT EXEMPLAR) Q10. Find the area bounded by the line y = x, X-axis and the lines x =-1to x =2. Q11. The area bounded by the curve y = f(x), the X-axis and x=1 and x =b is (b-1)sin Gb +4). Find f(x). @ Short Answer Type-i Questions 2 eit Q1. Find the area of the ellipse ~4Y 1 using a? BF integration method. (NCERY EXERCISE, NCERY EXEMPLAR) Q.2. Find the area of the ellipse x7 + 9y? =36 using integration. (CBSE 2020) Q3. Sketch the region bounded by the lines 2x + y =8, y =2, y =4and the Y-axis. Hence, obtain its area using integration. (CBSE 2023) Q4, Find the area of the region bounded by the line y =3x+2, X-axis and the ordinates x =-1 and xe t. (NCERT EXERCISE) 7) Short Answer Type-l! Questions o 2 Q1. Find the area bounded by the ellipse a 7 =1 a and the ordinates x =Oand x =ae. (NCERTEXERCISE) Q2. Find the area bounded by the curve y =cosx between x =Oand x =2n. Qa. Find the area of the region bounded by y =| x |, x $5 inthe first quadrant. Q4. Using integration, find the area of the region bounded by y =mx (m>0), x =1, x =2 and the X-axis. (CBSE 2023) Q6. Find the area of the region bounded by the parabola y ? = 4ax and the straight line y =mx. (NCERT EXERCISE) Q6. Find the area bounded by the lines y =|x -2|, x = 1,« =3and the X-axis. Q7. Find the area of the minor segment of the circle x? +y? =4 cut-off by the line x=1, using integration. (CBSE 2023) Q8. Find the area bounded by the curve y =cosx between x =Oand x = =. Q9. Find the area of the following region using integration: {(«, y) ty? S2x and y 2 x-4} . Long Answer Type Questions (CBSE 2023) Q1. Using integration, find the area of the region bounded by the parabola y? = 4ax and its latus rectum. (NCERT EXERCISE; CBSE 2023) Q2. Find the area of minor portion of the curve 9x? 416y*=144 intercepted by the line x =2 using definite integration. Qa. Using integration, find the area of the region in the first quadrant enclosed by the line x+y =2, the parabola y? = x and the X-axis. (NCERT; CBSE SQP 2022 Tarm-2) Q4, Using integration, find the area of the region in the first quadrant enclosed by the X-axis, the Line y = xand the circle x? +y? =32. (NCERT EXERCISE, CSE 2018) Q6. Using integration, find the area of the region bounded by the circle x? +y? = 16 line y = x and Y-axis, but lying in the 1st quadrant. (case 2023) Q6. Find the area of triangle whose two vertices formed from the X-axis and the line y =3-| x |. Q7. Find the ratio of the areas of two parts of the circle x? +y? =a’ divided by the line x =50. (NCERT EXEMPLAR) Q 8. Find the area of the regian bounded by the curve y =tanx,line x =—and the X-axis. 4 Q 9. Using integration, find the area of the region {Qx,y):O0 1 * OF 0.0) x TiP ” Do practice for taking limits of y. .. Required area of region AOBA 4 =2 «Area of shaded region =2f, x dy 22 W dy = tye= 22 sq. units 3 3 2. Let the line x =2 meets the parabola at pointsP and Q. TRICK Curve is symmetrical about X-axis je, area of both portions are equal numerically. ve We have to find the area of region POQP which Is double the area of shaded region PSOP. + Required area =2 x Area of shaded region =2[, y dx 22. BX dx =i B[ ar» WB] 28] oO By2 ay? =9F 2? 0) = 2 =2 sq. units 3. -. Required area = Area of shaded region. =| +1)dx] + fie +1)dx yor+t x= 2 = (3,4) (2,0) (1,0) 11 (-2,-1) ° @-TiP | [Learn to draw the graphs correctly. y 2 xe2y+3 1 yol dy x +5 a-4 De From the figure. required area of the shaded region A =f, Icos x |dx af « 2 fi cos x dx + Jjp(-c0s x) dx =[sin x] #? -[(sin x] 42 =[sinZ-sno] -[sinx-si 5| 2 2 = (1-0) -(O-1)=2 sq. units [EOMMONJERR()R Students fail to apply the limits correctly. 10. We have, y=x -. Required area = area of shaded region a =] J xdx|+ -l 2 Jxdx a 0 2 yal 2 lo x2 7 -3 41212245 zt sq. units *2 U. Given, f? F(x) dx =(b-1)sin (3b + 4) hr Area of function = 4 =(x-1)sin (3x +44) On differentiating, we get f(x) =sin(3x + 4) + 3(x-1)-cos (3x + 4) Short Answer Type-! Questions ey? 1. Equation of ellipse: rte (1) a For ellipse, major axis = 20, Minor axis = 2b and centre = (0, 0) (yTiP Learn to sketch the graphs of circle, ellipse and parabola froma standard equatian. y ~- Ellipse Is symmetrical about both axes. -. Required area = 4 xArea of shaded region a 7 4fs ‘ dx =4[52 be x? dx a 2140 Tx? +tatsin!X «| O12 pe a a 4b71 20 4 | <—| = In 1— ei ya sin ie) 4b | memati it: x= a? x= = nab 5q, units (From eq. (1)) 2. Given, x?49y?=36 or a6, ri => o*=36 > a=+6 and b?=4 => b=at2 Also, 9y* =36-x* > ys +136 -x? For first quadrant, y = 16 =x? y (0, 2) iad ». Required area = 4 x Area of shaded region 6] 2 = Afy den Af, 536 —x? dx = 3 [ere dx -3[ See = sin! (2) 0 -$[08 <£-0| 212m sq. units 3 ERROR Some students fail to find the standard equation of the ellipse and hence get wrong figure. De 3. Given equation of line Is 2x +y =8, which intercept X and Y-axes at points (4, 0) and (0, 8). ¥ \ 0.8) (4, 0) x +y=8 v 4 4/(8-y Area of bounced region = [* xdy = f° (a 1 zy eee =~(32-8-(16-2)) Pe 2 =3(24 -14)= = =5 sq. units 4, From the figure, the Une y =3x +2, meets X-axis at Ke Zand its graph below the X-axis for x € (-. -4) and above the X-axis for x ¢(-4.1} TRICK i f(x)dx =[. f(x) dx +ff f(x) dx, wherea y?-2y-8=0 = (y-4)(y+2)=0 = y=s4,-2 => X=B,2 -. Area of shaded region = [ (x2 ~x,)dy 4 2 “Llo-s-(5) 2 ay = Yay 2 6 |, andy = x-4/sa straight Une which Intersect X andY Long Answer Type Questions AL TA chord of a parabola passing through its focus and perpendicular to its axis is known as latus rectum of| parabola. Equation of parabola: y? = 4ax (1) and equation of latus rectum: x =a ..(2) Let the latus rectum meets the parabola at points P and Q. Y x’ ys ux y" We have to find the area of region POQSP which Is double the area of shaded region PSOP. Therefore required area = 2 x area of shaded region =2[py dx=2[) Vaax dx 3/27" =2 x20 [°x"? dx = 4fa| «2A (3c = | = 210. (0? -0}=£V0-aV6 8 2 ==0* 5q. units 3° 54 2. Equation of ellipse: 9x? +16y7=144 Al) 2 ye 2 ng AP iV simone 4 or —+2a1 form +551 9 [ ab? | For ellipse, semi-major axis =a=4 and minor axis =b=3 Equation of line: K=2 (2) Clearly, line is parallel to Y-axls. Let line (2), meets the ellipse (1) at point P and Q. Therefore, required area PAQRP = 2 x Area of shaded part =2f yd 92] 2 fgg 97x (from eq. (1)) 1 4 z 3 Apa = 53); V6 - x eas [ova ~x? dx De ©. a8 FS ° Put the et af x from eq. (2) into eq. (1). 7 y2+y? =32 s > 2y?=32 > y=t4 a ~ For the first quadrant, y = 4. a x q LY. (0.0)0 Ri(2, 0) /A (4, 0) and yed x =y Is a Une which passes through (0, 0) and (4, 4). Q . Required area = Area of OBL + Area of LBA v = “x Ix + “ff —x? dx mien =f, xdx+], 32 db 2 2 x 2 2 a, -1/ TRICK fvo -x die Na -x +0 (2)}+c | (ade ert! (e< 4 eri Nw [Sera | x? ‘ xVa2-x2 32 1 xX ia 2) 5s] 9/2842 24 oe * [ 2 I, “| 2 2" a |, =$[ asin fix V2 +a" 3] 2 x x =3[8 x5 -2N3-8 2] = 4-33 sq. units 3. The given line and parabola meet at the points (1, 1) and (4,2). -($-9} + [1 #1618160" +) =84+16-5-8-16-= 2 4 =On-4c= 40 5q. units Sometimes students takes wrong limits of common region and get the wrong value of area. eg., if wey dx (this iswrong) Divide the shaded region in two parts where the curves intersect and toke the right values of limits of both parts. 5, The area bounded by the circle x? + y? =16,y = x and :. Required area = Area of shaded region X-axis Is the area OABO. ={" VK dx+f (2-x) dx On solving x?+y?=16 and y=x, we have re 7 (y)P+y? =16 x2 x2 2 y2 =|——| +|2x-— = yorye sib 3/2 Jo 2, s 2y? =16 2 Y =2(1-0)+ (p2-2]-(2-3) w2 yg 3 2212-2 7 oe rts 3 2 6 4, x*+y* =32 Is a circle whose centre Is (0, 0) and x radius Is 4/2. (ee oy B . y= V32-x7 Now, x?#+y4=32 (I) and xey oi(2) Y Ma 8 => y? =B8 (4, 4) = y=2y2 z (In the first quadrant, y Is positive) x x ©) (1, i; (4 J, 0) When y = 22, then x =2¥2 way: So, the point of intersectlon of the given line and circle in the first quadrant Is (22,22). -. Required area = Area of the shaded region OABO ¥ = Area of OACO + Area of ABCA De When x =1, theny = V3. So, the point of intersection of the given line and circle in the first quadrant Is (1, V3) Required area = Area of the shaded region OABCO = Area of OACO + area of ABCA J ocx ace f? y dx xix ff feo? de 10. We have, (x, y):Osysx7+1LOsysx+10sxs2} Here consider y=x?41, which Is an equation of parabola open upwards with vertex (0, 1). Also, y = x + Uwhich Is an equation of line, intercept X and Y axes at -land 1 respectively. On solving above equations, x4laxtl = x?-x=0 = x=] > yel2 So, point of Intersections are (0, 1) and (1. 2). v yexel yex4l aa (1) £ x "| | y’ xol . Required area =|, ydx + JPydx = fy(x?+ 1) dea fi (x4 }) dx 2 Lane 2 -|5 ‘|, + [Ss al -[$+1-0]|$+2-(2+1)| 44 4-3 2812-9 Beg us 3 2 6 Tl. Given curves are x =O andx+2lyl=1 Now. x+2|yl=0 Wheny >0O, then x+2y =1 When y x?, the X-axis and the ordinate x =2. Q10. Draw the graph of the curve y =|sin x |and find the area bounded by the curve, X-axis and ordinates x=s-nto2z. Short Answer Type-II Questions Q11. Sketch the graph of y=|x+3] and evaluate Jo, Ler 31d Q12. Find the area bounded by the curve y =cotx, X-axis and the lines x= to x=—=. 2 4 Long Answer Type Questions Q13. Find the area bounded between the curve y? = 4x, line x+y =3and Y-axis. Q14. Find the area between X-axis, curve x = y and its normal y +2x =3 at the point (1, 1). De