Ex 7.7, 11 - Integrate root x2 - 8x + 7 dx - Class 12 - Ex 7.7

Ex 7.7, 11 - Chapter 7 Class 12 Integrals - Part 2


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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.