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Misc 4 - Sketch graph of y = |x+3| and evaluate integral 0 -> 6

Misc 4 - Chapter 8 Class 12 Application of Integrals - Part 2
Misc 4 - Chapter 8 Class 12 Application of Integrals - Part 3

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Misc 4 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’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)] +[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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.