![Area =
2
3
[Area of rectangle PQRS ]
© 2021, BYJU'S. All rights reserved
=
2
3
4 × 4
=
16
3
3, −3
1, −3
0
𝑋
𝑌
𝑦2
= 2𝑥 − 𝑥2
𝑦 = −3
2
1, 1
4
4](https://image.slidesharecdn.com/areaboundedbyinverseofthefunction1-230104113054-cbb73c3a/85/Area-Bounded-by-Inverse-of-The-Function-1-pdf-22-320.jpg)
Area Bounded by Inverse of The Function (1).pdf
- 1. Area Bounded by Inverse of the Function Area under Curve © 2021, BYJU'S. All rights reserved
- 2. Road Map © 2021, BYJU'S. All rights reserved A rea b ou n d ed by Inverse F u n c tion S ome Imp ortant Resu lts S h ift of O rig in
- 3. Area bounded by the function: 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1(i.e., ellipse) 𝑦2 𝑏2 = 1 − 𝑥2 𝑎2 ⇒ 𝑦2 = 𝑏2 𝑎2 𝑎2 − 𝑥2 𝐴 = 4 0 𝑎 𝑏 𝑎 𝑎2 − 𝑥2 𝑑𝑥 0 𝑋 𝑌 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 𝑎 𝑏 ⇒ 𝐴 = 𝜋𝑎𝑏 © 2021, BYJU'S. All rights reserved = 4𝑏 𝑎 𝑥 2 𝑎2 − 𝑥2 + 𝑎2 2 sin−1 𝑥 𝑎 0 𝑎 Result 1 = 4𝑏 𝑎 𝑎2 2 sin−1 𝑎 𝑎
- 4. Area bounded between 𝑦2 = 4𝑎𝑥 and 𝑥2 = 4𝑏𝑦:- 𝐴 = 0 4𝑎 2 3𝑏 1 3 2 𝑎 𝑥 − 𝑥2 4𝑏 𝑑𝑥 𝑦2 = 4𝑎𝑥 ⋯ 𝑖 𝑦 = 𝑥2 4𝑏 ⋯ 𝑖𝑖 © 2021, BYJU'S. All rights reserved Solving 𝑖 𝑎𝑛𝑑 (𝑖𝑖), 𝑥, 𝑦 ≡ 4𝑎 2 3𝑏 1 3, 4𝑎 1 3𝑏 2 3 4𝑎 2 3𝑏 1 3, 4𝑎 1 3𝑏 2 3 0 𝑋 𝑌 𝑦2 = 4𝑎𝑥 𝑥2 = 4𝑏𝑦 Result 2
- 5. © 2021, BYJU'S. All rights reserved 4𝑎 2 3𝑏 1 3, 4𝑎 1 3𝑏 2 3 0 𝑋 𝑌 𝑦2 = 4𝑎𝑥 𝑥2 = 4𝑏𝑦 ⇒ 𝐴 = 2 𝑎 𝑥 3 2 ⋅ 2 3 − 𝑥3 12𝑏 0 4𝑎 2 3𝑏 1 3 ⇒ 𝐴 = 16𝑎𝑏 3
- 6. 𝐴 = 16𝑎𝑏 3 ⇒ 𝐴 = 4𝑎 × 4𝑏 3 ⇒ 𝐴 = Coeff. of 𝑥 × Coeff. of 𝑦 3 © 2021, BYJU'S. All rights reserved Area bounded between 𝑦2 = 4𝑎𝑥 and 𝑥2 = 4𝑏𝑦:- Example Area bounded between 𝑦 = 𝑥 and 𝑥 = 𝑦 0 𝑋 𝑌 𝑦 = 𝑥 𝑥 = 𝑦 𝐴 = Coeff. of 𝑥 × (Coeff. of 𝑦) 3 𝐴 = 1 × 1 3 = 1 3 𝑥2 = 𝑦, 𝑦2 = 𝑥 Given, Point to remember
- 7. Area bounded between 𝑦2 = 4𝑎𝑥 and 𝑦 = 𝑚𝑥:- 0 𝑋 𝑌 𝑦2 = 4𝑎𝑥 𝑦 = 𝑚𝑥 𝑥 = 4𝑎 𝑚2 𝐴 = 0 4𝑎 𝑚2 2 𝑎 𝑥 − 𝑚𝑥 𝑑𝑥 𝑚2𝑥2 = 4𝑎𝑥 ⇒ 𝑥 𝑚2𝑥 − 4𝑎 = 0 ⇒ 𝑥 = 0, 𝑥 = 4𝑎 𝑚2 © 2021, BYJU'S. All rights reserved Result 3 Solving equations of curve, = 8𝑎2 3𝑚3 = 2 𝑎 2𝑥 3 2 3 − 𝑚𝑥2 2 0 4𝑎 𝑚2
- 8. Area bounded between 𝑦2 = 4𝑎𝑥(or any parabola) and its double ordinate (any line ⊥ to 𝑥-axis): 0 𝑋 𝑌 𝑦2 = 4𝑎𝑥 𝑃 𝑥, 𝑦 𝑄 𝑅 𝑆 𝑥 Double ordinate 𝑦 2𝑦 Shaded area = 2 3 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 𝑃𝑄𝑅𝑆 © 2021, BYJU'S. All rights reserved Result 4
- 9. Area enclosed by 𝑥2 16 + 𝑦2 9 = 1 ? S O L U T I O N Q U E S T I O N Area = 𝜋𝑎𝑏 © 2021, BYJU'S. All rights reserved Area = 𝜋 × 4 × 3 = 12𝜋 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 ⋯ ℎ𝑒𝑟𝑒 𝑎 = 4, 𝑏 = 3
- 10. If ellipse 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 meets 𝑥 − axis in 𝐴 and 𝑦 − axis in 𝐵 in 1𝑠𝑡 Quadrant, Find the Area bounded between Arc 𝐴𝐵 and chord 𝐴𝐵. S O L U T I O N Q U E S T I O N 0 𝑋 𝑌 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 𝑥 𝑎 + 𝑦 𝑏 = 1 Green Shaded area = = 𝜋𝑎𝑏 4 − ∆ = 𝜋𝑎𝑏 4 − 𝑎𝑏 2 0, 𝑏 𝑎, 0 © 2021, BYJU'S. All rights reserved Area of ellipse in 𝐼𝑠𝑡 quadrant – Area of triangle
- 11. Find the area bounded between 𝑥 = 𝑎 cos 𝑡 , 𝑦 = 𝑏 sin 𝑡. S O L U T I O N Q U E S T I O N cos 𝑡 = 𝑥 𝑎 , sin 𝑡 = 𝑦 𝑏 ⇒ sin2 𝑡 + cos2 𝑡 = 1 ⇒ 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 Area = 𝜋𝑎𝑏 © 2021, BYJU'S. All rights reserved 0 𝑋 𝑌 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 𝑎 𝑏 Converting polar form into Cartesian form
- 12. Find the area bounded between 4𝑥2 + 9𝑦2 = 36. S O L U T I O N Q U E S T I O N 𝑥2 9 + 𝑦2 4 = 1 Area = 𝜋 × 3 × 2 = 6𝜋 𝑎 = 3, 𝑏 = 2 Area = 𝜋𝑎𝑏 © 2021, BYJU'S. All rights reserved 0 𝑋 𝑌 𝑥2 9 + 𝑦2 4 = 1 3 2
- 13. Find the area bounded between 𝑥 = 𝑎 sin3 𝑡 , 𝑦 = 𝑎 cos3 𝑡. S O L U T I O N Q U E S T I O N 𝑥 2 3 + 𝑦 2 3 = 𝑎 2 3 0, 𝑎 𝑎, 0 −𝑎, 0 0, −𝑎 ∴ sin2 𝑡 + cos2 𝑡 = 1 ⇒ 𝑥 𝑎 2 3 + 𝑦 𝑎 2 3 = 1 ⇒ 𝑥 2 3 + 𝑦 2 3 = 𝑎 2 3 Area = 4 0 𝑎 𝑦𝑑𝑥 sin 𝑡 = 𝑥 𝑎 1 3 , cos 𝑡 = 𝑦 𝑎 1 3 © 2021, BYJU'S. All rights reserved Converting polar form into Cartesian form = 4 0 𝑎 𝑎 2 3 − 𝑥 2 3 3 2 𝑑𝑥 ⋯ (𝑖)
- 14. 𝑥 2 3 + 𝑦 2 3 = 𝑎 2 3 0, 𝑎 𝑎, 0 −𝑎, 0 0, −𝑎 Area = 4 0 𝜋 2 𝑎 cos3 𝑡 × 3𝑎 sin2 𝑡 cos 𝑡 𝑑𝑡 = 12𝑎2 0 𝜋 2 cos4 𝑡 ⋅ sin2 𝑡 𝑑𝑡 = 12𝑎2 × 3 ⋅ 1 × 1 6 ⋅ 4 ⋅ 2 × 𝜋 2 © 2021, BYJU'S. All rights reserved = 3𝑎2𝜋 8 ⋯ 𝑓𝑟𝑜𝑚 (𝑖) ⇒ 𝑑𝑥 = 3𝑎 sin2 𝑡 cos 𝑡 𝑑𝑡 𝑥 = 𝑎 sin3 𝑡 U.L, if 𝑥 = 𝑎, 𝑡 = 𝜋 2 L.L, if 𝑥 = 0, 𝑡 = 0
- 15. Find the area bounded between 𝑥2 + 𝑦2 = 1 and 𝑦 + 𝑥 = 1 S O L U T I O N Q U E S T I O N Green Shaded Area 𝑋 𝑌 𝑥2 + 𝑦2 = 1 1 1 0 𝑦 + 𝑥 = 1 © 2021, BYJU'S. All rights reserved = 4 × 𝜋 1 2 4 − 1 2 × 1 × 1 = 𝜋 − 2 = 4 × (Area of Circle − Area of Square) = 𝜋 1 2 − 4 × 1 2 ⋯ 4 × (1𝑠𝑡 quadrant)
- 16. Q U E S T I O N Find the area bounded between 𝑥2 + 𝑦2 = 𝜋2 & 𝑦 = sin 𝑥 in −𝜋, 𝜋 S O L U T I O N 0 𝑋 𝑌 𝑦 = sin 𝑥 Shaded area = Area of circle 2 = 𝜋 × 𝜋2 2 = 𝜋3 2 𝜋, 0 −𝜋, 0 © 2021, BYJU'S. All rights reserved 𝑥2 + 𝑦2 = 𝜋2 ⋯ (∵ sin 𝑥 𝑖𝑠 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛) ⋯ (∵ 𝑟𝑎𝑑𝑖𝑢𝑠 = 𝜋)
- 17. Find the area bounded between 𝑥 − 2 2 + 𝑦 − 3 2 = 32 and line 𝑦 = 𝑥 + 1 S O L U T I O N Q U E S T I O N (2, 3) By observation, 𝑦 = 𝑥 + 1 passes through centre ⇒ 𝐴 = 𝜋 × 32 2 = 16𝜋 Circle with center (−ℎ, −𝑘) ≡ 2, 3 , 𝑟2 = 32 𝑦 = 𝑥 + 1 © 2021, BYJU'S. All rights reserved Comparing with 𝑥 − ℎ 2 + 𝑦 − 𝑘 2 = 𝑟2, Shaded area = Area of circle 2
- 18. Parabola 𝑦2 = 4𝑥 and 𝑥2 = 4𝑦 divides the square region bounded by lines 𝑥 = 4, 𝑦 = 4 and co-ordinate axis in 𝑆1, 𝑆2, 𝑆3, then 𝑆1 ∶ 𝑆1 ∶ 𝑆3 = ? S O L U T I O N Q U E S T I O N 𝑆2 = Coeff. of 𝑥 × Coeff. of 𝑦 3 ⇒ 𝑆2 = 4 × 4 3 = 16 3 ⋯ (𝑖) 0 𝑋 𝑌 𝑦2 = 4𝑥 𝑥2 = 4𝑦 𝑆2 𝑆1 𝑆3 4, 4 𝑥 = 4 𝑦 = 4 𝑆1 + 𝑆2 + 𝑆3 = 16 ⋯ (𝑖𝑖) © 2021, BYJU'S. All rights reserved Solving 𝑦2 = 4𝑥 and 𝑥2 = 4𝑦, 𝑥, 𝑦 ≡ (4, 4) From diagram, ⋯ 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒 = 42
- 19. 0 𝑋 𝑌 𝑦2 = 4𝑥 𝑥2 = 4𝑦 𝑆2 𝑆1 𝑆3 4, 4 𝑥 = 4 𝑦 = 4 𝑆1 + 𝑆3 = 16 − 16 3 = 32 3 ⇒ 2𝑆1 = 32 3 ∴ 𝑆1 ∶ 𝑆2 ∶ 𝑆3 = 1 ∶ 1 ∶ 1 © 2021, BYJU'S. All rights reserved ⇒ 𝑆1 = 𝑆3 ⋯ 𝑓𝑟𝑜𝑚 𝑖 𝑎𝑛𝑑 (𝑖𝑖) ⋯ 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑤. 𝑟. 𝑡 𝑦 = 𝑥 ⇒ 𝑆1 = 16 3 ⋯ (𝑖𝑖𝑖) ⋯ 𝑓𝑟𝑜𝑚 𝑖 , 𝑖𝑖 , (𝑖𝑖𝑖)
- 20. If the area bounded between 𝑥 = 𝑎𝑦2 and 𝑦 = 𝑎𝑥2 is 1 , then 𝑎 = ? S O L U T I O N Q U E S T I O N 𝑦2 = 𝑥 𝑎 & 𝑥2 = 𝑦 𝑎 Area = Coeff. of 𝑥 × Coeff. of 𝑦 3 = 1 𝑎 × 1 𝑎 3 = 1 3𝑎2 = 1 3𝑎2 = 1 ⇒ 𝑎2 = 1 3 ⇒ 𝑎 = ± 1 3 © 2021, BYJU'S. All rights reserved Given, ⋯ 𝑔𝑖𝑣𝑒𝑛 0 𝑋 𝑌 𝑎𝑦2 = 𝑥 𝑎𝑥2 = 𝑦 1 𝑎 , 1 𝑎
- 21. Area bounded between 𝑦 = 2𝑥 − 𝑥2 and 𝑦 + 3 = 0. S O L U T I O N Q U E S T I O N 𝑦 = 2𝑥 − 𝑥2 , 𝑎 < 0 ⇒ downward parabola 𝑦 = −𝑥 𝑥 − 2 ⇒ Roots = 0, 2 Solving line 𝑦 = −3 and the curve 𝑦 = 2𝑥 − 𝑥2 −3 = 2𝑥 − 𝑥2 ⇒ 𝑥2 − 2𝑥 − 3 = 0 ⇒ 𝑥 − 3 𝑥 + 1 = 0 3, −3 −1, −3 0 𝑋 𝑌 𝑦2 = 2𝑥 − 𝑥2 𝑦 = −3 2 1, 1 4 4 © 2021, BYJU'S. All rights reserved 𝑃 𝑄 𝑅 𝑆 ⇒ 𝑥 = 3, −1
- 22. Area = 2 3 [Area of rectangle PQRS ] © 2021, BYJU'S. All rights reserved = 2 3 4 × 4 = 16 3 3, −3 1, −3 0 𝑋 𝑌 𝑦2 = 2𝑥 − 𝑥2 𝑦 = −3 2 1, 1 4 4
- 23. In case co-ordinate axes are shifted parallelly then the area of figure and slope line do not change. Shifting of origin:- © 2021, BYJU'S. All rights reserved Shifting of Origin
- 24. Q U E S T I O N 𝐶1: 𝑦2 − 2𝑦 + 4𝑥 + 5 = 0 𝐶2: 𝑥2 + 2𝑥 − 𝑦 + 2 = 0 Find the area bounded between: © 2021, BYJU'S. All rights reserved S O L U T I O N 𝐶1: 𝑦2 − 2𝑦 + 4𝑥 + 5 = 0 𝐶2: 𝑥2 + 2𝑥 − 𝑦 + 2 = 0 𝑦2 − 2𝑦 + 1 = −4 𝑥 + 1 𝑦 − 1 2 = −4 𝑥 + 1 𝑌2 = −4𝑋 𝑥2 + 2𝑥 + 1 = 𝑦 − 1 𝑥 + 1 2 = 𝑦 − 1 𝑋2 = 𝑌
- 25. ⇒ 𝐴 = Coeff. of 𝑥 × Coeff. of 𝑦 3 0 𝑋 𝑌 𝑌2 = −4𝑋 𝑋2 = 𝑌 ⇒ 𝐴 = −4 × 1 3 = 4 3 sq. units © 2021, BYJU'S. All rights reserved
- 26. Q U E S T I O N Find the area bounded between the smaller area bounded by ellipse 9𝑥2 + 4𝑦2 − 36𝑥 + 8𝑦 + 4 = 0 and line 3𝑥 + 2𝑦 = 10. Ellipse: S O L U T I O N ⇒ 3𝑥 − 6 2 + 2𝑦 + 2 2 = 36 ⇒ 9 𝑥 − 2 2 + 4 𝑦 + 1 2 = 36 9𝑥2 − 36𝑥 + 36 + 4𝑦2 + 8𝑦 + 4 = 36 ⇒ 𝑥−2 2 4 + 𝑦+1 2 9 = 1 ⇒ 𝑋2 4 + 𝑌2 9 = 1 𝑎 = 2 , 𝑏 = 3; 𝑎 < 𝑏 © 2021, BYJU'S. All rights reserved 2 3 0 𝑋 𝑌 𝑋 2 + 𝑌 3 = 1 𝑋2 4 + 𝑌2 9 = 1
- 27. Line: 3𝑥 − 6 + 2𝑦 + 2 = 4 + 2 ⇒ 3 𝑥 − 2 + 2 𝑦 + 1 = 6 3𝑋 + 2𝑌 = 6 ⇒ 𝑋 2 + 𝑌 3 = 1 © 2021, BYJU'S. All rights reserved Area = Area of ellipse – Area of triangle = 3𝜋 2 − 3 2 3 0 𝑋 𝑌 𝑋 2 + 𝑌 3 = 1 𝑋2 4 + 𝑌2 9 = 1 = 𝜋⋅2⋅3 4 − 3 = 𝜋𝑎𝑏 4 − 1 2 × 3 × 2
- 28. 𝜋 4 3𝜋 4 5𝜋 4 −𝜋 4 −3𝜋 4 −7𝜋 4 −5𝜋 4 7𝜋 4 The area enclosed by the curves 𝑦 = sin 𝑥 + cos 𝑥 and 𝑦 = cos 𝑥 − sin 𝑥 over the interval 0, 𝜋 2 is : Q U E S T I O N 1 2 𝑦 = sin 𝑥 + cos 𝑥 𝑦 = cos 𝑥 − sin 𝑥 0 sin 𝑥 + cos 𝑥 cos 𝑥 − sin 𝑥 JEE Adv 2013 S O L U T I O N © 2021, BYJU'S. All rights reserved 𝑋 𝑌 = 2 cos 𝑥 − 𝜋 4 = 2 cos 𝑥 ⋅ 1 2 + sin 𝑥 ⋅ 1 2 = 2 cos 𝑥 + 𝜋 4 = 2 cos 𝑥 ⋅ 1 2 − sin 𝑥 ⋅ 1 2
- 29. 𝜋 4 3𝜋 4 5𝜋 4 −𝜋 4 −3𝜋 4 −7𝜋 4 −5𝜋 4 7𝜋 4 𝑦 = cos 𝑥 − sin 𝑥 𝑦 = sin 𝑥 + cos 𝑥 0 𝜋 2 1 2 © 2021, BYJU'S. All rights reserved
- 30. Area = 2 0 𝜋 4 sin 𝑥 𝑑𝑥 + 2 𝜋 4 𝜋 2 cos 𝑥 𝑑𝑥 = 2 2( 2 − 1) = 0 𝜋 4 (sin 𝑥 + cos 𝑥) − (cos 𝑥 − sin 𝑥) 𝑑𝑥 + 𝜋 4 𝜋 2 (sin 𝑥 + cos 𝑥) − (sin 𝑥 − cos 𝑥) 𝑑𝑥 𝜋 4 0 𝜋 2 𝑦 = cos 𝑥 − sin 𝑥 𝑦 = sin 𝑥 + cos 𝑥 1 2 𝑦 = sin 𝑥 − cos 𝑥 𝑦 = cos 𝑥 − sin 𝑥 © 2021, BYJU'S. All rights reserved 𝑋 𝑌
- 31. Area Bounded by Inverse of Function If 𝑓 𝑥 and inverse of 𝑓 𝑥 are always mirror images of each other. Area bounded between 𝑦 = 𝑓−1 𝑥 , 𝑥 = 𝑎, 𝑥 = 𝑏, 𝑥-axis = Area bounded between 𝑦 = 𝑓 𝑥 , 𝑦 = 𝑎, 𝑦 = 𝑏, 𝑦-axis 𝑏 𝑎 𝑦 = 𝑓 𝑥 𝑦 = 𝑓−1 𝑥 𝑋 𝑌 0 𝑎 𝑏 © 2021, BYJU'S. All rights reserved
- 32. SOLUTION QUESTION 𝑂 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 © 2021, BYJU'S. All rights reserved
- 33. 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝑓(𝑥) ⋯ ∵ 𝑓 𝑥 𝑖𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑦 © 2021, BYJU'S. All rights reserved
- 34. 𝑖 Area of 𝑓(𝑥) with 𝑦-axis 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝑓(𝑥) = Area of 𝑦 with 𝑥-axis ⋯ ∵ 𝑓 𝑥 𝑖𝑠 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑦 Area ⋯ (𝑖) 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝑓(𝑥) © 2021, BYJU'S. All rights reserved = 0 𝜋 𝑦 𝑑𝑥
- 35. 𝑖𝑖 Area of 𝑓(𝑥) with 𝑥-axis 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝑓(𝑥) = Area of 𝑔(𝑥) with 𝑦-axis = Area of square − Area of 𝑓(𝑥) with 𝑦-axis 𝜋 𝑂 𝜋 𝑌 𝑋 𝑦 = 𝑥 + sin 𝑥 𝑦 = 𝑥 𝑓(𝑥) 𝜋2 2 + 2 ⋯ 𝑓𝑟𝑜𝑚 (𝑖) © 2021, BYJU'S. All rights reserved
- 36. Let 𝑓: −1,2 → [0, ∞) be a continuous function such that 𝑓 𝑥 = 𝑓 1 − 𝑥 ∀ 𝑥 ∈ −1,2 . Let 𝑅1 = −1 2 𝑥𝑓 𝑥 𝑑𝑥 , and 𝑅2 be the area of the region bounded by 𝑦 = 𝑓 𝑥 , 𝑥 = −1, 𝑥 = 2 and 𝑥−axis . Then Q U E S T I O N A B C D 𝑅1 = 2𝑅2 𝑅1 = 3𝑅2 3𝑅1 = 𝑅2 2𝑅1 = 𝑅2 IIT - JEE 2011 𝑅1 𝑅1 = −1 2 𝑥𝑓 𝑥 𝑑𝑥 ⋯ (𝑖) = −1 2 (1 − 𝑥)𝑓 1 − 𝑥 𝑑𝑥 ⋯ 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝑎 𝑏 𝑓 𝑎 + 𝑏 − 𝑥 𝑑𝑥 S O L U T I O N © 2021, BYJU'S. All rights reserved
- 37. Adding, 𝑖 𝑎𝑛𝑑 (𝑖𝑖) 2𝑅1 = −1 2 𝑓 𝑥 𝑑𝑥 = 𝑅2 ⋯ 𝑅2 = −1 2 𝑓 𝑥 𝑑𝑥 ∴ Option 𝑪 is correct option. 𝑅1 = −1 2 𝑓(𝑥) − 𝑥𝑓(𝑥) 𝑑𝑥 ⋯ (𝑖𝑖) 𝑅1 = −1 2 (1 − 𝑥)𝑓(𝑥) 𝑑𝑥 ⋯ 𝑓 𝑥 = 𝑓 1 − 𝑥 © 2021, BYJU'S. All rights reserved
- 38. Q U E S T I O N A B C D 8 3 2 − 1 2 8 3 2 − 1 4 3 2 − 1 2 4 3 2 + 1 The area (in sq. units ) of the region 𝐴 = 𝑥, 𝑦 : 𝑥 − 1 𝑥 ≤ 𝑦 ≤ 2 𝑥, 0 ≤ 𝑥 < 2 , where 𝑡 denotes the greatest integer function, is : JEE Main Sept 2019 S O L U T I O N 0 ≤ 𝑦 ≤ 2 𝑥; 𝑥 − 1 ≤ 𝑦 ≤ 2 𝑥; 0 ≤ 𝑥 < 1 1 ≤ 𝑥 < 2 © 2021, BYJU'S. All rights reserved
- 39. Area = Area under 𝑦 = 2 𝑥 - Area under 𝑦 = 𝑥 − 1 = 4 3 𝑥 3 2 0 2 − 1 2 = 0 2 2 𝑥𝑑𝑥 − 1 2 ⋅ 1 ⋅ 1 = 8 3 2 − 1 2 0 𝑦 = 𝑥 − 1 (2, 2) 𝑌 𝑋 1 2 𝑦 = 2 𝑥 © 2021, BYJU'S. All rights reserved
- 40. The area (in sq. units ) of the region 𝐴 = 𝑥, 𝑦 : 𝑥 + 𝑦 ≤ 1 , 2𝑦2 ≥ |𝑥| , is : Q U E S T I O N A B C D 5 6 1 3 7 6 1 6 JEE Main Sept 2020 𝑌 𝑋 0 𝑥 + 𝑦 = 1 2𝑦2 = 𝑥 1 2 , 1 2 Solving, 2𝑦2 = 𝑥 𝑎𝑛𝑑 𝑥 + 𝑦 = 1 ⇒ 𝑦 = −1, 1 2 S O L U T I O N ⇒ (𝑥, 𝑦) ≡ 1 2 , 1 2 ⋯ 𝑖𝑛 1𝑠𝑡 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 © 2021, BYJU'S. All rights reserved
- 41. Area of shaded region = 4 𝑥 − 𝑥2 2 − 1 2 ⋅ 2 3 𝑥 3 2 0 1 2 = 4 3 8 − 1 6 𝑌 𝑋 0 𝑥 + 𝑦 = 1 2𝑦2 = 𝑥 1 2 , 1 2 = 4 0 1 2 1 − 𝑥 − 𝑥 2 𝑑𝑥 = 4 1 2 − 1 8 − 2 3 ⋅ 1 2 2 = 10 12 = 5 6 © 2021, BYJU'S. All rights reserved
- 42. Summary Sheet © 2021, BYJU'S. All rights reserved Area bounded by inverse of function If 𝑓 𝑥 and inverse of 𝑓 𝑥 are always mirror images of each other. Area bounded between 𝑦 = 𝑓−1 𝑥 , 𝑥 = 𝑎, 𝑥 = 𝑏, 𝑥-axis = Area bounded between 𝑦 = 𝑓 𝑥 , 𝑦 = 𝑎, 𝑦 = 𝑏, 𝑦-axis 𝑏 𝑎 𝑦 = 𝑓 𝑥 𝑦 = 𝑓−1 𝑥 𝑋 𝑌 0 𝑎 𝑏