Misc 4 - Sketch graph of y = |x + 3| and integral 0 to -6 - Area between curve and curve

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  1. Chapter 8 Class 12 Application of Integrals
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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+πŸ‘| y = |π‘₯+3| = {β–ˆ(π‘₯+3 π‘“π‘œπ‘Ÿ π‘₯+3β‰₯0@βˆ’(π‘₯+3) π‘“π‘œπ‘Ÿ π‘₯+3<0 )─ = {β–ˆ(π‘₯+3 π‘“π‘œπ‘Ÿ π‘₯β‰₯βˆ’3@βˆ’(π‘₯+3) π‘“π‘œπ‘Ÿ π‘₯+3<βˆ’3 )─ Finding ∫_(βˆ’πŸ”)^πŸŽβ–’γ€–β”‚π’™+πŸ‘β”‚ 𝒅𝒙〗 ∫_(βˆ’6)^0β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— =∫_(βˆ’6)^(βˆ’3)β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— βˆ’βˆ«_(βˆ’3)^0β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— =∫_(βˆ’6)^(βˆ’3)β–’γ€–βˆ’(π‘₯+3) 𝑑π‘₯ +∫_(βˆ’3)^0β–’γ€–(π‘₯+3) 𝑑π‘₯γ€—γ€— =[βˆ’π‘₯^2/2βˆ’3π‘₯]_(βˆ’6)^(βˆ’3) +[π‘₯^2/2+3π‘₯]_(βˆ’3)^( 0) = (βˆ’(βˆ’3)^2)/( 2)βˆ’3 Γ— (βˆ’3)βˆ’[(βˆ’(βˆ’6)^2)/( 2)βˆ’3(βˆ’6)] +[βˆ’0^2/2+3 Γ—0]βˆ’[(βˆ’3)^2/2+3 Γ— (βˆ’3)] = (βˆ’9)/( 2)βˆ’(βˆ’9)βˆ’[(βˆ’36)/( 2)βˆ’(βˆ’18)]βˆ’[9/2βˆ’9] =(βˆ’9)/( 2)+9+0βˆ’9/( 2)+9 =βˆ’9+18 =9

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.