Question 3 - Miscellaneous - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Miscellaneous
Misc 1
Important
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Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Question 3 Evaluate β«_0^1βπ^(2 β3π₯)β‘ππ₯ as a limit of a sum . I=β«_0^1βπ^(2 β3π₯)β‘ππ₯ I=β«_0^1βγπ^2 . π^(β3π₯)γβ‘ππ₯ I=π^2 β«_0^1βπ^(β3π₯)β‘ππ₯ Solving I1 separately β«_0^1βπ^(β3π₯) ππ₯ Putting π = 0 π =1 β = (π β π)/π = (1 β 0)/π = 1/π π(π₯)=π^(β3π₯) We know that β«1_π^πβγπ₯ ππ₯γ =(πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) Hence we can write β«_0^1βπ^(β3π₯) ππ₯ =(1β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π^(β3π₯) π(0)=π^(β3(0))=1 π(β)=π^(β3β) π(2β)=π^(β3(2β))=π^(β6β) π((πβ1)β)=π^(β3(πβ1)β) Hence, our equation becomes β«_0^1βπ^(β3π₯) ππ₯ =limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π(πβ1)β) = limβ¬(nββ) 1/π (1+π^(β3β)+π^(β6β)+ β¦β¦+π^(β3(π β 1) β) ) Let S = 1+π^(β3β)+π^(β6β)+ β¦β¦+π^(β3(π β 1) β) It is a G.P. with common ratio (r) r = π^(β3β)/1 = π^(β3β) We know Sum of G.P = a((π^π β 1)/(π β 1)) Replacing a by 1 and r by π^(β3β) , we get S = 1(((π^(β3β) )^π β 1)/(π^(β3β) β 1))= (π^(β3πβ) β 1)/(π^(β3β) β 1) Thus β΄ β«_0^1βπ^(β3π₯) ππ₯ =limβ¬(nββ) 1/π (1+π^(β3β)+π^(β6β)+ β¦+π^(β3(π β 1) β) ) Putting the value of S, we get =limβ¬(nββ) 1/π ((π^(β3πβ) β 1)/(π^(β3β) β 1)) = (πππ)β¬(πββ) 1/π ((π^(β3πβ) β 1)/(β3β . (π^(β3β) β 1)/(β3β))) = (πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ) . 1/( (π^(β3β) β 1)/(β3β)) = (πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ) . (πππ)β¬(πββ) 1/( (π^(β3β) β 1)/(β3β)) Solving (π₯π’π¦)β¬(π§ββ) ( π)/(( π^(βππ) β π)/(βππ)) As nββ β 1/β ββ β β β0 β΄ limβ¬(nββ) ( 1)/(( π^(β3β) β 1)/(β3β)) = limβ¬(hβ0) ( 1)/(( π^(β3β) β 1)/(β3β)) = 1/1 = 1 Thus, our equation becomes β«1_0^1βγπ^(β3π₯) ππ₯γ =(πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ).(πππ)β¬(πββ) 1/( (π^(β3β) β 1)/(β3β)) = (πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ). 1 = (πππ)β¬(πββ) (π^(β3π . 1/π) β 1)/(β3π (1/π) ) = (πππ)β¬(πββ) (π^(β3) β 1)/(β3) = 1/1 = 1 Thus, our equation becomes β«1_0^1βγπ^(β3π₯) ππ₯γ =(πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ).(πππ)β¬(πββ) 1/( (π^(β3β) β 1)/(β3β)) = (πππ)β¬(πββ) (π^(β3πβ) β 1)/(β3πβ). 1 = (πππ)β¬(πββ) (π^(β3π . 1/π) β 1)/(β3π (1/π) ) = (πππ)β¬(πββ) (π^(β3) β 1)/(β3) = (π^(β3) β 1)/(β3) = (1 β π^(β3))/3 = (1 β 1/π^3 )/3 = (π^3 β 1)/(3π^3 ) Putting the values of I1 in (1) I=π^2Γ1/3 [(π^3 β 1)/π^3 ] I1=1/3 [(π^3 β 1)/π] π=π/π [π^πβ π/π]
