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, 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
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.