Misc 2 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 16, 2024 by

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Misc 2 Sketch the graph of y = |π‘₯+3| and evaluate ∫_(βˆ’6)^0β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— Let’s 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 )─ Now, Required Area = ∫_(βˆ’πŸ”)^πŸŽβ–’γ€–β”‚π’™+πŸ‘β”‚ 𝒅𝒙〗 =∫_(βˆ’6)^(βˆ’3)β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— βˆ’βˆ«_(βˆ’3)^0β–’γ€–β”‚π‘₯+3β”‚ 𝑑π‘₯γ€— =∫_(βˆ’πŸ”)^(βˆ’πŸ‘)β–’γ€–βˆ’(𝒙+πŸ‘) 𝒅𝒙 +∫_(βˆ’πŸ‘)^πŸŽβ–’γ€–(𝒙+πŸ‘) 𝒅𝒙〗〗 =[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)] +[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)]+[0]βˆ’[9/2βˆ’9] =(βˆ’9)/2+9+0βˆ’9/2+9 =βˆ’9+18 =πŸ— square units

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo