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Ex 7.7, 7 - Integrate root 1 + 3x - x2 - Chapter 7 CBSE - Ex 7.7

Ex 7.7, 7 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.7, 7 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.7, 7 √(1+3π‘₯βˆ’π‘₯2) ∫1β–’γ€–βˆš(1+3π‘₯βˆ’π‘₯^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’(βˆ’3π‘₯+π‘₯^2 ) ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’(π‘₯^2βˆ’3π‘₯) ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’[π‘₯^2βˆ’2(3/2)(π‘₯)] ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’[π‘₯^2βˆ’2(3/2)(π‘₯)+(3/2)^2βˆ’(3/2)^2 ] ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’[(π‘₯βˆ’3/2)^2βˆ’9/4] ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1βˆ’(π‘₯βˆ’3/2)^2+9/4) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(1+γ€–9/4βˆ’(π‘₯βˆ’3/2)γ€—^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(γ€–13/4βˆ’(π‘₯βˆ’3/2)γ€—^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(γ€–(√13/2)^2βˆ’(π‘₯βˆ’3/2)γ€—^2 ) 𝑑π‘₯γ€— =(π‘₯ βˆ’ 3/2)/2 √((√13/2)^2βˆ’(π‘₯βˆ’3/2)^2 )+(√13/2)^2/2 𝑠𝑖𝑛^(βˆ’1) ((π‘₯ βˆ’ 3/2))/(√13/2)+𝐢 It is of the form ∫1β–’γ€–βˆš(π‘Ž^2βˆ’π‘₯^2 ) 𝑑π‘₯=1/2 π‘₯√(π‘Ž^2βˆ’π‘₯^2 )+π‘Ž^2/2 𝑠𝑖𝑛^(βˆ’1) π‘₯/π‘Ž+𝐢〗 ∴ Replacing π‘₯ by π‘₯βˆ’3/2 and a by √13/2 , we get =((2π‘₯ βˆ’ 3)/2)/2 √(13/4βˆ’(π‘₯βˆ’3/2)^2 )+(13/4)/2 𝑠𝑖𝑛^(βˆ’1) (((2π‘₯ βˆ’ 3)/2))/(√13/2)+𝐢 =(2π‘₯ βˆ’ 3)/4 √(13/4βˆ’(π‘₯βˆ’3/2)^2 )+13/8 𝑠𝑖𝑛^(βˆ’1) (2π‘₯ βˆ’ 3)/√13+𝐢 =(2π‘₯ βˆ’ 3)/4 √(13/4βˆ’[π‘₯^2+9/4βˆ’2(π‘₯)(3/2)] )+13/8 𝑠𝑖𝑛^(βˆ’1) (2π‘₯ βˆ’ 3)/√13+𝐢 =(2π‘₯ βˆ’ 3)/4 √(13/4βˆ’π‘₯^2βˆ’9/4+3π‘₯)+13/8 𝑠𝑖𝑛^(βˆ’1) (2π‘₯ βˆ’ 3)/√13+𝐢 =(2π‘₯ βˆ’ 3)/4 √(13/4βˆ’9/4+3π‘₯βˆ’π‘₯^2 )+13/8 𝑠𝑖𝑛^(βˆ’1) (2π‘₯ βˆ’ 3)/√13+𝐢 =(πŸπ’™ βˆ’ πŸ‘)/πŸ’ √(𝟏+πŸ‘π’™βˆ’π’™^𝟐 )+πŸπŸ‘/πŸ– π’”π’Šπ’^(βˆ’πŸ) ((πŸπ’™ βˆ’ πŸ‘)/βˆšπŸπŸ‘)+π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.