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Ex 7.10, 19 Show that โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ) ๐‘” (๐‘ฅ) ๐‘‘๐‘ฅ=2โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ) ๐‘‘๐‘ฅ, if f and g are defined as ๐‘“(๐‘ฅ)=๐‘“(๐‘Žโˆ’๐‘ฅ) and ๐‘”(๐‘ฅ)+๐‘”(๐‘Žโˆ’๐‘ฅ)=4 Let I =โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ) ๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ I =โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ) [4โˆ’๐‘”(๐‘Žโˆ’๐‘ฅ)] ๐‘‘๐‘ฅ I = โˆซ_0^๐‘Žโ–’[4.๐‘“(๐‘ฅ)โˆ’๐‘“(๐‘ฅ)๐‘”(๐‘Žโˆ’๐‘ฅ)] ๐‘‘๐‘ฅ I = 4โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘”(๐‘Žโˆ’๐‘ฅ) ใ€—ใ€— ๐‘‘๐‘ฅ I = 4โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘Žโˆ’๐‘ฅ) ๐‘”(๐‘Žโˆ’(๐‘Žโˆ’๐‘ฅ)) ใ€—ใ€— ๐‘‘๐‘ฅ I = 4โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘”(๐‘ฅ) ใ€—ใ€— ๐‘‘๐‘ฅ I =4โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ)๐‘‘๐‘ฅโˆ’Iใ€— I +I=4โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ 2I=4โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ I=2โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ โˆด โˆซ_0^๐‘Žโ–’ใ€–๐‘“(๐‘ฅ) ๐‘”(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=2โˆซ_0^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ Hence Proved

  1. Chapter 7 Class 12 Integrals
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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo