# Example 14 - Chapter 8 Class 12 Application of Integrals

Last updated at Nov. 15, 2019 by Teachoo

Last updated at Nov. 15, 2019 by Teachoo

Transcript

Example 14 Prove that the curves 𝑦2=4𝑥 and 𝑥2=4𝑦 divide the area of the square bounded by 𝑥=0, 𝑥=4, 𝑦=4 and 𝑦=0 into three equal parts Step 1: Drawing figure 𝑦2=4𝑥 𝑥2=4𝑦 Square To Prove : Area OPQA = Area OAQB = Area OBQR Area OPQA Area OPQA = 04𝑦 𝑑𝑥 Here, 4𝑦= 𝑥2 𝑦= 𝑥24 So, Area OPQA = 04 𝑥24𝑑𝑥 = 14 𝑥3304 = 112× 43− 03 = 112 × 64−0 = 6412= 163 Area OBQR Since Area is on 𝑦−𝑎𝑥𝑖𝑠 , we use formula 𝑥 𝑑𝑦 Area OBQR = 04𝑥 𝑑𝑦 Here, 𝑦2=4𝑥 𝑥= 𝑦24 So, Area OBQR = 04 𝑦24𝑑𝑦 = 14 𝑦3304= 112× 43− 03= 112 × 64−0= 163 Area OAQB Area OAQB = Area OBQP – Area OAQP Finding Area OBQP Area OBQP = 04𝑦 𝑑𝑥 Here, 𝑦2=4𝑥 𝑦=± 4𝑥 As OBQP is in 1st quadrant, we use positive 𝑦= 4𝑥 Area OBQP = 04 4𝑥 𝑑𝑥 =2 04 𝑥 𝑑𝑥 =2 04 𝑥 12 𝑑𝑥 =2 × 𝑥 12+1 12+104 =2 × 𝑥 32 3204 =2 × 23 𝑥 3204 = 43 4 32− 0 32 = 43 8−0 = 323 Area OAQP Area OAQP = 04𝑦 𝑑𝑥 Here, 𝑥2=4𝑦 𝑦= 𝑥24 Area OAQP = 04 𝑥24𝑑𝑥 = 14 𝑥2+12+104 = 14 ×3 𝑥304 = 112 43− 03 = 112 × 64−0 = 163 ∴ Area OAQB = Area OBQD – Area OAQP = 323− 163 = 163 So, Area OAQB = Area OAQP = Area OBRQ = 163 Hence Proved.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.