Area between curve and curve

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Misc 4 Sketch the graph of y = |π₯+3| and evaluate β«_(β6)^0βγβπ₯+3β ππ₯γ Letβs 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€ Now, Required Area = β«_(βπ)^πβγβπ+πβ 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)] +[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)]+[0]β[9/2β9] =(β9)/2+9+0β9/2+9 =β9+18 =π square units