Definite Integral as a limit of a sum

Chapter 7 Class 12 Integrals
Concept wise

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Question 2 Evaluate β«_0^2βπ^π₯ ππ₯ as the limit of a sum . β«_0^2βπ^π₯ ππ₯ Putting π = 0 π = 2 β = (π β π)/π = (2 β 0)/π = 2/π π(π₯)=π^π₯ We know that β«1_π^πβγπ₯ ππ₯γ =(πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) Hence we can write β«_0^2βπ^π₯ ππ₯ =(2β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =2 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π^π₯ π(0)=π^0=1 π(0+β)=π^(0 + β)=π^β π(0+2β)=π^(0 + 2β)=π^2β π(0+(πβ1)β)=π^(0 + (πβ1)β)=π^(πβ1)β Hence, our equation becomes β΄ β«_0^2βπ^π₯ ππ₯ =2 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π(πβ1)β) = 2 .limβ¬(nββ) 1/π (1+π^β+π^2β+ β¦β¦+π^((π β 1) β) ) Let S = π^β+π^2β+ β¦β¦+π^((π β 1) β) It is a G.P. with common ratio (r) r = π^2β/π^β = π^(2β β β) = π^β Sum of G.P. S = (π (1 β π^π ))/(1 β π) = (π^β (1 β (π^β )^π ))/(1 β π^β ) = (π^β (1 β π^πβ ))/(1 β π^β ) = (2β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦β¦+π(0+(πβ1)β) = 2 .limβ¬(nββ) 1/π (1+π^β+π^2β+ β¦β¦+π^((π β 1) β) ) Let S = π^β+π^2β+ β¦β¦+π^((π β 1) β) It is a G.P. with common ratio (r) r = π^2β/π^β = π^(2β β β) = π^β Sum of G.P. S = (π (1 β π^π ))/(1 β π) = (π^β (1 β (π^β )^π ))/(1 β π^β ) = (π^β (1 β π^πβ ))/(1 β π^β ) Thus β΄ β«_0^2βπ^π₯ ππ₯ = 2 .limβ¬(nββ) 1/π (1+π^β+π^2β+ β¦β¦+π^((π β 1) β) ) Putting the value of S, we get = 2 .limβ¬(nββ) 1/π (1+ (π^β (1 β π^πβ ))/(1 β π^β )) = 2 (limβ¬(nββ) 1/π +limβ¬(nββ) 1/π (π^β ((1 β π^πβ)/(1 β π^β )))) "Taking β1 common from" "numerator and denominator " = 2 (1/β +limβ¬(nββ) (π^β/π . ( π^πβ β 1)/(π^β β 1))) "Multiplying and dividing denominator by h" = 2 (0+limβ¬(nββ) (π^β/π . ( π^πβ β 1)/(β . ( π^β β 1)/β))) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β)(( 1)/(( π^β β 1)/β)) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . limβ¬(nββ) (( 1)/(( π^β β 1)/β)) Solving (π₯π’π¦)β¬(π§ββ) ( π)/(( π^π β π)/π) As nββ β 2/β ββ β β β0 β΄ limβ¬(nββ) ( 1)/(( π^β β 1)/β) = limβ¬(hβ0) ( 1)/(( π^β β 1)/β) = 1/1 = 1 (ππ πππ limβ¬(t β 0) ( π^π‘ β 1)/π‘=1) Thus, our equation becomes β΄ β«_0^2βπ^π₯ ππ₯ = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . limβ¬(nββ) (( 1)/(( π^β β 1)/β)) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . 1 = 2 . limβ¬(nββ) π^( 2/π)/π (( π^(π . 2/π) β 1)/(2/π)) = 2 . limβ¬(nββ) γ πγ^( 2/π) (( π^2 β 1)/2) = 2 (π^( 2/β) ((π^2 β 1))/2) (ππ πππ β= 2/π) = 2 . π^0 ((π^2 β 1))/2 = 2/2 . 1 (π^2β1) = 1 . 1 (π^2β1) = π^πβπ