Example 25 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at Dec. 23, 2019 by Teachoo
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Example 25 Important Deleted for CBSE Board 2023 Exams You are here
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Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 25 Find β«_0^2β(π₯^2+1) ππ₯ as the limit of a sum . β«_0^2β(π₯^2+1) ππ₯ Putting π = 0 π = 2 β = (π β π)/π = (2 β 0)/π = 2/π π(π₯)=π₯^2+1 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«_0^2β(π₯^2+1) ππ₯ =(2β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =2 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π₯^2+1 π(0)=0^2+1=0+1=1 π(β)=β^2+1=(2/π)^2+1=4/π^2 +1 π(2β)=(2β)^2+1=γ4βγ^2+1=4(2/π)^2+1=16/π^2 +1 β¦.. π(πβ1)β=((πβ1)β)^2+1=γ(πβ1)^2 (2/π)γ^2+1 =(πβ1)^2 Γ 4/π^2 +1 Hence, our equation becomes = 2 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π(πβ1)β) = 2 limβ¬(nββ) 1/π (1+(4/π^2 +1)+(16/π^2 +1" " )+ β¦β¦+((4(π β 1)^2)/π^2 +1)) = 2 limβ¬(nββ) 1/π ((1 + 1 + 1β¦π π‘ππππ )+0+ 4/π^2 +16/π^2 + β¦(4(π β 1)^2)/π^2 ) = 2 limβ¬(nββ) 1/π (π +0+ 4/π^2 +16/π^2 + β¦β¦(4(π β 1)^2)/π^2 ) = 2 limβ¬(nββ) 1/π (π+ 4/π^2 (1+4+ β¦β¦+(π β 1)^2 ) ) = 2 limβ¬(nββ) 1/π (π+ 4/π^2 (1^2+2^2+ β¦β¦β¦+(π β 1)^2 ) ) = 2 limβ¬(nββ) 1/π (π+ 4/π^2 ((π β 1) π(2π β 1))/6) = 2 limβ¬(nββ) 1/π (π+ 4/π ((π β 1) (2π β 1))/6) = 2 limβ¬(nββ) 1/π (π+ 2/3π (πβ1) (2πβ1)) = 2 limβ¬(nββ) (π/π + 2/(3π^2 ) (πβ1) (2πβ1)) We know that 1^2+2^2+ β¦β¦+π^2= (π(π + 1) (2π +1))/6 1^2+2^2+ β¦β¦+(πβ1)^2= ((π β 1)(π β 1+ 1) (2(π β1)+1))/6 = ((π β 1) π(2π β 1))/6 = 2 limβ¬(nββ) (π/π + 2/3 ((π β 1))/π ((2π β 1))/π) = 2 limβ¬(nββ) (1+ 2/3 (1β 1/π) (2β 1/π)) = 2 (1+ 2/3 (1β0) (2β0)) = 2 (1+ 2/3 Γ2) = 2 (1+ 4/3) = 2 Γ 7/3 = ππ/π (limβ¬(nββ) 1/π=0" " )
