Check sibling questions

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

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.