Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.7, 11 ∫1β–’γ€–βˆš(π‘₯2 βˆ’8π‘₯+7) " " γ€— 𝑑π‘₯ 1/2 (x - 4)√(π‘₯2 βˆ’8π‘₯+7) + 9 log |π‘₯βˆ’4+√(π‘₯2βˆ’8π‘₯+7)| + C 1/2 (x + 4)√(π‘₯2 βˆ’8π‘₯+7) + 9 log |π‘₯+4+√(π‘₯2βˆ’8π‘₯+7)| + C 1/2 (x - 4)√(π‘₯2 βˆ’8π‘₯+7) - 3√2 log |π‘₯βˆ’4+√(π‘₯2βˆ’8π‘₯+7)| + C 1/2 (x - 4)√(π‘₯2 βˆ’8π‘₯+7) + 9/2 log |π‘₯βˆ’4+√(π‘₯2βˆ’8π‘₯+7)| + C ∫1β–’γ€–βˆš(π‘₯^2βˆ’8π‘₯+7) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(π‘₯^2βˆ’2(4)(π‘₯)+7) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(π‘₯^2βˆ’2(4)(π‘₯) γ€–+(4)γ€—^2βˆ’(4)^2+7) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš((π‘₯βˆ’4)^2βˆ’16+7) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš((π‘₯βˆ’4)^2βˆ’9 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš((π‘₯βˆ’4)^2βˆ’(3)^2 ) 𝑑π‘₯γ€— =(π‘₯ βˆ’ 4)/2 √((π‘₯βˆ’4)^2βˆ’(3)^2 )βˆ’(3)^2/2 π‘™π‘œπ‘”|π‘₯βˆ’4+√((π‘₯βˆ’4)^2βˆ’(3)^2 )|+𝐢 =(π‘₯ βˆ’ 4)/2 √(π‘₯^2βˆ’8π‘₯+16βˆ’9 )βˆ’9/2 π‘™π‘œπ‘”|π‘₯βˆ’4+√(π‘₯^2βˆ’8π‘₯+16βˆ’9)|+𝐢 =(π‘₯ βˆ’ 4)/2 √(π‘₯^2βˆ’8π‘₯+7)βˆ’9/2 π‘™π‘œπ‘”|π‘₯βˆ’4+√(π‘₯^2βˆ’8π‘₯+7)|+𝐢 ∴ Option (D) is correct. It is of the form ∫1β–’γ€–βˆš(π‘₯^2βˆ’π‘Ž^2 ) 𝑑π‘₯=π‘₯/2 √(π‘₯^2βˆ’π‘Ž^2 )βˆ’π‘Ž^2/2 π‘™π‘œπ‘”|π‘₯+√(π‘₯^2βˆ’π‘Ž^2 )|+𝐢〗 ∴ Replacing π‘₯ by π‘₯βˆ’4 and a by 3 , we get

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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.