1. Chapter 8 Class 12 Application of Integrals
2. Serial order wise
3. Miscellaneous

Transcript

Misc 4 - Chapter 8 Class 12 Application of Integrals - NCERT Solution Sketch the graph of y = |x+3| and evaluate β« -6 -> 0 |x + 3| dx Draw the graph y = |π+π| y = |π₯+3| = {β(π₯+3 πππ π₯+3β₯0@β(π₯+3) πππ π₯+3<0 )β€ = {β(π₯+3 πππ π₯β₯β3@β(π₯+3) πππ π₯+3<β3 )β€ Finding β«_(βπ)^πβγβπ+πβ ππγ β«_(β6)^0βγβπ₯+3β ππ₯γ =β«_(β6)^(β3)βγβπ₯+3β ππ₯γ ββ«_(β3)^0βγβπ₯+3β ππ₯γ =β«_(β6)^(β3)βγβ(π₯+3) ππ₯ +β«_(β3)^0βγ(π₯+3) ππ₯γγ =[βπ₯^2/2β3π₯]_(β6)^(β3) +[π₯^2/2+3π₯]_(β3)^( 0) = (β(β3)^2)/( 2)β3 Γ (β3)β[(β(β6)^2)/( 2)β3(β6)] +[β0^2/2+3 Γ0]β[(β3)^2/2+3 Γ (β3)] = (β9)/( 2)β(β9)β[(β36)/( 2)β(β18)]β[9/2β9] =(β9)/( 2)+9+0β9/( 2)+9 =β9+18 =9

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