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Areas between Two Curves (in Hindi)
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This lesson explains the method to to calculate areas between two curves

## Vineet Loomba is teaching live on Unacademy Plus

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5/8 ....done mam
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sir ..1 st q me hm direct y p element leke aur phir uska integration nhi kr skte kya..?
MA
I have passed 12th in 2014 I have given only one exam for B tech entrance test I am 24years old now so can I attept 2020 B tech entrance test please tell me soon Thanks alot
MA
I have passed 12th in 2014 I have given only one exam for B tech entrance test I am 24years old now so can I attept 2020 B tech entrance test please tell me soon Thanks alot
1. 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)

2. 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!!

3. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The area bounded by the curves y = r and r + = 2 above X-axis is : SOLUTION: (D) Let us first find the points ofintersection of curves. Solving y =x, and r -2 simultaneously, we get (x2-1)(x2 + 2) = 0 1 ) 2 4 3 3 2 A=(-1,0) and B=(1,1) sq. units. Shaded Area= +1W2-2-x2)dr MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

4. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Compute the area of the fiqure bounded by the straight lines Solution: y=2, (x-1)2 =-(y-1) The required area =12(y1-y2)dx x 0, x-2 and the curves y-2", y-2x-x2 R(2,4) where y, 2 and ya-2x - x2 (2 -2x x2)dx 2* In 2 2 (0,1 3 8 3n2 n2 3 M2,0) sq.units In 2 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

5. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration: Find the area bounded by the curve y (x 1) (x 2) (x 3) lying between the ordinates x 0 and x 3 and x-axis Solution To determine the sign, we follow the usual rule of change of sign. for x > 3 for 2 <x <3 for 1 < x 2 for x <1 y +ve y F-ve 3 Now let F(x)-|(x-1)(x-2)(x-3)dx =J(x3-6x2 + 11x _ 6) dx = x4-2x3 +_xs_6x. 4 2 , F(2)2, F(3) 4 Hence required Area IF) FO)+F2) F()-F3) F2) 2sq.units. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

6. 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)

7. 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)