Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.