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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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Transcript

Ex 7.7, 7 √(1+3𝑥−𝑥2) ∫1▒〖√(1+3𝑥−𝑥^2 ) 𝑑𝑥〗 =∫1▒〖√(1−(−3𝑥+𝑥^2 ) ) 𝑑𝑥〗 =∫1▒〖√(1−(𝑥^2−3𝑥) ) 𝑑𝑥〗 =∫1▒〖√(1−[𝑥^2−2(3/2)(𝑥)] ) 𝑑𝑥〗 =∫1▒〖√(1−[𝑥^2−2(3/2)(𝑥)+(3/2)^2−(3/2)^2 ] ) 𝑑𝑥〗 =∫1▒〖√(1−[(𝑥−3/2)^2−9/4] ) 𝑑𝑥〗 =∫1▒〖√(1−(𝑥−3/2)^2+9/4) 𝑑𝑥〗 =∫1▒〖√(1+〖9/4−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =∫1▒〖√(〖13/4−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =∫1▒〖√(〖(√13/2)^2−(𝑥−3/2)〗^2 ) 𝑑𝑥〗 =(𝑥 − 3/2)/2 √((√13/2)^2−(𝑥−3/2)^2 )+(√13/2)^2/2 𝑠𝑖𝑛^(−1) ((𝑥 − 3/2))/(√13/2)+𝐶 It is of the form ∫1▒〖√(𝑎^2−𝑥^2 ) 𝑑𝑥=1/2 𝑥√(𝑎^2−𝑥^2 )+𝑎^2/2 𝑠𝑖𝑛^(−1) 𝑥/𝑎+𝐶〗 ∴ Replacing 𝑥 by 𝑥−3/2 and a by √13/2 , we get =((2𝑥 − 3)/2)/2 √(13/4−(𝑥−3/2)^2 )+(13/4)/2 𝑠𝑖𝑛^(−1) (((2𝑥 − 3)/2))/(√13/2)+𝐶 =(2𝑥 − 3)/4 √(13/4−(𝑥−3/2)^2 )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−[𝑥^2+9/4−2(𝑥)(3/2)] )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−𝑥^2−9/4+3𝑥)+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(2𝑥 − 3)/4 √(13/4−9/4+3𝑥−𝑥^2 )+13/8 𝑠𝑖𝑛^(−1) (2𝑥 − 3)/√13+𝐶 =(𝟐𝒙 − 𝟑)/𝟒 √(𝟏+𝟑𝒙−𝒙^𝟐 )+𝟏𝟑/𝟖 𝒔𝒊𝒏^(−𝟏) ((𝟐𝒙 − 𝟑)/√𝟏𝟑)+𝑪

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.