Misc 4 - Chapter 8 Class 12 Application of Integrals

Transcript

Misc 4 - Chapter 8 Class 12 Application of Integrals - NCERT Solution Sketch the graph of y = |x+3| and evaluate ∫ -6 -> 0 |x + 3| dx Draw the graph y = |𝒙+𝟑| y = |𝑥+3| = {█(𝑥+3 𝑓𝑜𝑟 𝑥+3≥0@−(𝑥+3) 𝑓𝑜𝑟 𝑥+3<0 )┤ = {█(𝑥+3 𝑓𝑜𝑟 𝑥≥−3@−(𝑥+3) 𝑓𝑜𝑟 𝑥+3<−3 )┤ Finding ∫_(−𝟔)^𝟎▒〖│𝒙+𝟑│ 𝒅𝒙〗 ∫_(−6)^0▒〖│𝑥+3│ 𝑑𝑥〗 =∫_(−6)^(−3)▒〖│𝑥+3│ 𝑑𝑥〗 −∫_(−3)^0▒〖│𝑥+3│ 𝑑𝑥〗 =∫_(−6)^(−3)▒〖−(𝑥+3) 𝑑𝑥 +∫_(−3)^0▒〖(𝑥+3) 𝑑𝑥〗〗 =[−𝑥^2/2−3𝑥]_(−6)^(−3) +[𝑥^2/2+3𝑥]_(−3)^( 0) = (−(−3)^2)/( 2)−3 × (−3)−[(−(−6)^2)/( 2)−3(−6)] +[−0^2/2+3 ×0]−[(−3)^2/2+3 × (−3)] = (−9)/( 2)−(−9)−[(−36)/( 2)−(−18)]−[9/2−9] =(−9)/( 2)+9+0−9/( 2)+9 =−9+18 =9