Example 28 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Definite Integration by properties - P2
Definite Integration by properties - P2
Last updated at April 16, 2024 by Teachoo
Example 28 Evaluate β«_(β1)^2β|π₯^3βπ₯| ππ₯ |π₯^3βπ₯|=|π₯(π₯^2β1)| =|π₯| |π₯^2β1| =|π₯| |(π₯β1)(π₯+1)| =|π₯| |π₯β1||π₯+1| Thus, π₯=0, π₯=1 & π₯=β1 β΄ |π₯^3βπ₯|= {β((βπ₯)(β(π₯β1))(β(π₯+1)) ππ π₯<β1@(βπ₯)"(β(π₯β1))" (π₯+1) ππ β1β€π₯<0@(π₯)"(β(π₯β1))" (π₯+1) ππ 0β€π₯<1@(π₯)(π₯β1)(π₯+1) ππ π₯β₯1)β€ |π₯^3βπ₯|= {β(βπ₯(π₯β1)(π₯+1) ππ π₯<β1@π₯(π₯β1)(π₯+1) ππ β1β€π₯<0@βπ₯(π₯β1)(π₯+1) ππ 0β€π₯<1@π₯(π₯β1)(π₯+1) ππ π₯β₯1)β€ |π₯^3βπ₯|= {β(β(π₯^3βπ₯) ππ π₯<β1@(π₯^3βπ₯) ππ β1β€π₯<0@β(π₯^3βπ₯) ππ 0β€π₯<1@(π₯^3βπ₯) ππ π₯β₯1)β€ Thus, our integration becomes β«_(β1)^2β|π₯^3βπ₯| ππ₯ =β«_(β1)^0β(π₯^3βπ₯) ππ₯ββ«_0^1β(π₯^3βπ₯) ππ₯+β«_1^2β(π₯^3βπ₯) ππ₯ =β«_(β1)^0βπ₯^3 ππ₯ββ«_(β1)^0βπ₯ ππ₯ββ«_0^1βπ₯^3 ππ₯+β«_0^1βπ₯ ππ₯+β«_1^2βπ₯^3 ππ₯ββ«_1^2βπ₯ ππ₯ =[π₯^4/4]_(β1)^0β[π₯^2/2]_(β1)^0β[π₯^4/4]_0^1+[π₯^2/2]_0^1+[π₯^4/4]_1^2β[π₯^2/2]_1^2 =[(0 β (β1)^4)/4]β[(0 β (β1)^2)/2]β[((1)^4 β 0)/4]+[((1)^2 β 0)/2]+[((2)^4 β (1)^4)/4]β[((2)^2 β (1)^2)/2] =(β1)/4 β [(β1)/2] β 1/4 + 1/2 + [(16 β 1)/4]β[(4 β 1)/2] =(β1)/4 + 1/2 β 1/4 + 1/2 +15/4 β 3/2 =(β1 + 2 β 1 + 2 + 15 β 6)/4 =ππ/π