Get live Maths 1-on-1 Classs - Class 6 to 12
Example 26 - Chapter 7 Class 12 Integrals (Term 2)
Last updated at March 16, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii)
Example 2 (i)
Example 2 (ii)
Example 2 (iii) Important
Example 3 (i)
Example 3 (ii) Important
Example 3 (iii)
Example 4
Example 5 (i)
Example 5 (ii)
Example 5 (iii) Important
Example 5 (iv) Important
Example 6 (i)
Example 6 (ii) Important
Example 6 (iii) Important
Example 7 (i)
Example 7 (ii) Important
Example 7 (iii)
Example 8 (i)
Example 8 (ii) Important
Example 9 (i)
Example 9 (ii) Important
Example 9 (iii) Important
Example 10 (i)
Example 10 (ii) Important
Example 11
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Example 23
Example 24
Example 25 Important Deleted for CBSE Board 2023 Exams
Example 26 Deleted for CBSE Board 2023 Exams You are here
Example 27 (i)
Example 27 (ii) Important
Example 27 (iii)
Example 27 (iv) Important
Example 28
Example 29
Example 30 Important
Example 31
Example 32
Example 33
Example 34 Important
Example 35 Important
Example 36 Important
Example 37
Example 38 Important
Example 39 Important
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 25 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Chapter 7 Class 12 Integrals
Serial order wise
Transcript
Example 26 Evaluate β«_0^2βπ^π₯ ππ₯ as the limit of a sum . β«_0^2βπ^π₯ ππ₯ 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^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«_0^2βπ^π₯ 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^π₯ π(0)=π^0=1 π(0+β)=π^(0 + β)=π^β π(0+2β)=π^(0 + 2β)=π^2β π(0+(πβ1)β)=π^(0 + (πβ1)β)=π^(πβ1)β Hence, our equation becomes β΄ β«_0^2βπ^π₯ ππ₯ =2 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π(πβ1)β) = 2 .limβ¬(nββ) 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^πβ ))/(1 β π^β ) = (2β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦β¦+π(0+(πβ1)β) = 2 .limβ¬(nββ) 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^πβ ))/(1 β π^β ) Thus β΄ β«_0^2βπ^π₯ ππ₯ = 2 .limβ¬(nββ) 1/π (1+π^β+π^2β+ β¦β¦+π^((π β 1) β) ) Putting the value of S, we get = 2 .limβ¬(nββ) 1/π (1+ (π^β (1 β π^πβ ))/(1 β π^β )) = 2 (limβ¬(nββ) 1/π +limβ¬(nββ) 1/π (π^β ((1 β π^πβ)/(1 β π^β )))) "Taking β1 common from" "numerator and denominator " = 2 (1/β +limβ¬(nββ) (π^β/π . ( π^πβ β 1)/(π^β β 1))) "Multiplying and dividing denominator by h" = 2 (0+limβ¬(nββ) (π^β/π . ( π^πβ β 1)/(β . ( π^β β 1)/β))) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β)(( 1)/(( π^β β 1)/β)) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . limβ¬(nββ) (( 1)/(( π^β β 1)/β)) Solving (π₯π’π¦)β¬(π§ββ) ( π)/(( π^π β π)/π) As nββ β 2/β ββ β β β0 β΄ limβ¬(nββ) ( 1)/(( π^β β 1)/β) = limβ¬(hβ0) ( 1)/(( π^β β 1)/β) = 1/1 = 1 (ππ πππ limβ¬(t β 0) ( π^π‘ β 1)/π‘=1) Thus, our equation becomes β΄ β«_0^2βπ^π₯ ππ₯ = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . limβ¬(nββ) (( 1)/(( π^β β 1)/β)) = 2 . limβ¬(nββ) π^β/π (( π^πβ β 1)/β) . 1 = 2 . limβ¬(nββ) π^( 2/π)/π (( π^(π . 2/π) β 1)/(2/π)) = 2 . limβ¬(nββ) γ πγ^( 2/π) (( π^2 β 1)/2) = 2 (π^( 2/β) ((π^2 β 1))/2) (ππ πππ β= 2/π) = 2 . π^0 ((π^2 β 1))/2 = 2/2 . 1 (π^2β1) = 1 . 1 (π^2β1) = π^πβπ
