## Vineet Loomba is teaching live on Unacademy Plus

APPLICATIONS OF INTEGRALS (AREA UNDER CURVES) JEE MAIN AND ADVANCED IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The area bounded by y x 4 and x y 2 is: 75 100 125 150 (B) 6 (C) 6 SOLUTION: (C) After drawing the figure, let us find the points of in- tersection of 125 =2x5--(4-9)--(8+27) + 4(5)-- x=-3,2 As(-3,5) and Bs(2,0) Shaded area, = | | (2-x)-(x2-4) | dr MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example : Find area bounded by y = 4-x, x-axis and the lines x = 0 and x = 2. Sol: By using the formula of Area Bounded by the x axis, we can4 obtain Required Area. 2 0 0 2 8 16 = | 4x--| = 8--=-sq. units 0 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find the area of the region enclosed by y = sin x, y = cos x and x-axis, 0 sx . 2 ea Sol: Find point of intersection is P. Therefore after obtaining the co-ordinates of P and then integrating with appropriate limits, we can obtain required Area. Hence, P is Required area 4 2 t/2 = 2--/2 sq. units At point of intersection P x =-as ordinates of y = sin x and ; y cos x are equal 4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration Using the method of integration, find the area of the region bounded by lines: 2x y 4, 3x- 2y =6and x-3y + 5 = 0 (JEE ADVANCED) Given equation of the lines are 2x + y = 4 3x-2y = 6 x- 3y5 0 Solving (i) and (i), we get (2, 0) Solving (i) and (ii), we get (4, 3) Solving) and (ii), we get (1, 2) (iii) (4,3) 4(3x-6 .: Required AreaX+5 2,0) ! (8 +20)--+5||-[(8-4)-(4-1)]-124-24)-(6-12)] -sq. units. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Enroll 934 ENROL for this Course (Free) Recommend Lessons (Like) E Rate and Review the Course ** 15 68 153 ratings - 30 reviews .Comments . Sharing with friends Share MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)