Ex 7.7, 7 - Chapter 7 Class 12 Integrals (Term 2)

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