Area as a sum

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Question 6 β«1_0^4β(π₯+π2π₯)ππ₯ Let I = β«1_0^4β(π₯+π2π₯)ππ₯ I = β«1_0^4βγπ₯ . ππ₯γ+β«1_0^4βγ π2π₯ . ππ₯γ Solving I1 and I2 separately Solving I1 β«1_0^4βγπ₯ ππ₯γ Putting π =0 π =4 β=(π β π)/π =(4 β 0)/π =4/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«1_0^4βγπ₯ ππ₯γ =(4β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =4 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)+β¦ +π((πβ1)β) Here, π(π₯)=π₯ π(0)=0 π(β)=β π (2β)=2β π((πβ1)β)=(πβ1)β Hence, our equation becomes β΄ β«_0^4βπ₯ ππ₯ =4 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)+β¦ +π((πβ1)β) = 4 (πππ)β¬(πββ) 1/π (0+β+2β+ β¦β¦+(πβ1)β) = 4 (πππ)β¬(πββ) 1/π ( β (1+2+ β¦β¦β¦+(πβ1))) We know that 1+2+3+ β¦β¦+π= (π (π + 1))/2 1+2+3+ β¦β¦+πβ1= ((π β 1) (π β 1 + 1))/2 = (π (π β 1) )/2 = 4 (πππ)β¬(πββ) 1/π ( (β . π(π β 1))/2) = 4 (πππ)β¬(πββ) ( π(π β 1)β/2π) = 4 (πππ)β¬(πββ) ( (π β 1)β/2) = 4 (πππ)β¬(πββ) ( (π β 1)4/(2 . π)) = 4 (πππ)β¬(πββ) ( 2(π/π β 1/π)) = 4 (πππ)β¬(πββ) ( 2(1β 1/π)) = 4( 2(1β 1/β)) [ππ πππ β=4/π] = 4( 2(1β0)) = 4Γ2 = π Solving I2 β«_0^4βπ^2π₯ ππ₯ Putting π = 0 π =4 β = (π β π)/π = (4 β 0)/π = 4/π π(π₯)=π^2π₯ 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«_0^4βπ^2π₯ ππ₯ =(4β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =4 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π^2π₯ π(0)=π^(2(0))=1 π(β)=π^2β π(2β)=π^(2(2β))=π^4β π((πβ1)β)=π^2(πβ1)β Hence we can write β«_0^4βπ^2π₯ ππ₯ =(4β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =4 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π^2π₯ π(0)=π^(2(0))=1 π(β)=π^2β π(2β)=π^(2(2β))=π^4β π((πβ1)β)=π^2(πβ1)β Hence, our equation becomes β΄ β«_0^4βπ^2π₯ ππ₯ =4 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π(πβ1)β) = 4 .limβ¬(nββ) 1/π (1+π^2β+π^4β+ β¦β¦+π^(2(π β 1) β) ) Let S = 1+π^2β+π^4β+ β¦β¦+π^(2(π β 1) β) It is a G.P. with common ratio (r) r = π^2β/1 = π^2β We know Sum of G.P = a((π^π β 1)/(π β 1)) Replacing a by 1 and r by π^2β , we get S = 1(((π^2β )^π β 1)/(π^2β β 1))= (π^2πβ β 1)/(π^2β β 1) Thus β΄ β«_0^4βπ^π₯ ππ₯ = 4 limβ¬(nββ) 1/π (1+π^2β+π^4β+ β¦β¦+π^(2(π β 1) β) ) Putting the value of S, we get = 4 .limβ¬(nββ) 1/π ((π^2πβ β 1)/(π^2β β 1)) = 4 (πππ)β¬(πββ) 1/π ((π^2πβ β 1)/(2β . (π^2β β 1)/2β)) = 4 (πππ)β¬(πββ) (π^2πβ β 1)/2πβ . 1/( (π^2β β 1)/2β) = 4 (πππ)β¬(πββ) (π^2πβ β 1)/2πβ . (πππ)β¬(πββ) 1/( (π^2β β 1)/2β) Solving (π₯π’π¦)β¬(π§ββ) ( π)/(( π^ππ β π)/ππ) As nββ β 2/β ββ β β β0 β΄ limβ¬(nββ) ( 1)/(( π^2β β 1)/2β) = limβ¬(hβ0) ( 1)/(( π^2β β 1)/2β) = 1/1 = 1 Thus, our equation becomes β«1_0^4βγππ₯ ππ₯γ ="4" (πππ)β¬(πββ) (π^2πβ β 1)/2πβ . (πππ)β¬(πββ) 1/( (π^2β β 1)/2β) " " = "4" (πππ)β¬(πββ) (π^2πβ β 1)/2πβ . 1 = 4 (πππ)β¬(πββ) (π^(2π . 4/π) β 1)/(2π (4/π) ) = 4 (πππ)β¬(πββ) (π^8 β 1)/8 (ππ πππ (πππ)β¬(π‘β0) (π^π‘ β 1)/π‘ =1) (ππ πππ β=4/π) = 4 (π^8 β 1)/8 = (π^π β π)/π Putting the values of I1 and I2 in I β΄ I = β«1_0^4βγπ₯ . ππ₯γ+β«1_0^4βγ π2π₯ . ππ₯γ = 8 + (π^8 β 1)/2 = (16 + π^8 β 1)/2 = (ππ + π^π)/π