Misc 2 - Chapter 8 Class 12 Application of Integrals

Last updated at April 16, 2024 by

Slide5.JPG

Slide6.JPG
Slide7.JPG

Go Ad-free

Transcript

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

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