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Ex 7.11
Ex 7.11, 2
Ex 7.11, 3 Important
Ex 7.11, 4
Ex 7.11, 5 Important
Ex 7.11, 6
Ex 7.11,7 Important
Ex 7.11,8 Important
Ex 7.11, 9
Ex 7.11, 10 Important
Ex 7.11, 11 Important
Ex 7.11, 12 Important
Ex 7.11, 13
Ex 7.11, 14
Ex 7.11, 15
Ex 7.11, 16 Important
Ex 7.11, 17
Ex 7.11, 18 Important You are here
Ex 7.11, 19
Ex 7.11, 20 (MCQ) Important
Ex 7.11, 21 (MCQ) Important
Last updated at Dec. 20, 2019 by Teachoo
Ex 7.11, 18 By using the properties of definite integrals, evaluate the integrals : β«_0^4β|π₯β1| ππ₯ |π₯β1|= {β( (π₯β1) ππ π₯β1β₯0@β(π₯β1) ππ π₯β1<0)β€ = {β((π₯β1,) ππ π₯β₯1@β(π₯β1) ππ π₯<1)β€ β΄ β«_0^4β|π₯β1|ππ₯=β«_0^1β|π₯β1|ππ₯+β«_1^4β|π₯β1|ππ₯ Using the property, P2 P2 :- β«_π^πβγπ(π₯)ππ₯=γ β«_π^πβγπ(π₯)ππ₯+β«_π^πβπ(π₯)ππ₯γ =β«_0^1βγβ(π₯β1)ππ₯+γ β«_1^4β(π₯β1)ππ₯ =β«_0^1βγ(βπ₯+1)ππ₯+γ β«_1^4β(π₯β1)ππ₯ =β«_0^1βγβπ₯ ππ₯+γ β«_0^1βγ1. ππ₯+β«_1^4βγπ₯ . ππ₯ββ«_1^4βγ1.ππ₯γγγ =β[π₯^2/2]_0^1+[π₯]_0^1β[π₯^2/2]_1^4β[π₯]_1^4 =β[((1)^2 β 0)/2]+[1β0]+[((4)^2β(1)^2)/2]β[4β1] =β1/2+1+[(16 β 1)/2]β3 =β1/2+15/2β3+1 =(14 )/2β2= 7 β 2 = 5