Area as a sum

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Question 5 β«1_(β1)^1βγππ₯ ππ₯γ β«1_(β1)^1βγππ₯ ππ₯γ Putting π =β1 π =1 β=(π β π)/π =(1 β (β1))/π =(1 + 1)/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«1_(β1)^1βγππ₯ ππ₯γ =(1 β(β1)) (πππ)β¬(πββ) 1/π (π(β1)+π(β1+β)+π(β1+2β)+ β¦+π(β1+(πβ1)β)) =2 (πππ)β¬(πββ) 1/π (π(β1)+π(β1+β)+π(β1+2β)+ β¦+π(β1+(πβ1)β)) Here, π(π₯)=π^π₯ π(β1)=π^(β1) π(β1+β)=π^(β1 + β) =π^(β1). π^β π (β1+2β)=π^(β1 + 2β)=π^(β1). π^2β π(β1+(πβ1)β)=π^(β1 + (π β 1)β)=π^(β1).π^(π β 1)β Hence, our equation becomes β«1_(β1)^1βγππ₯ ππ₯γ =2 (πππ)β¬(πββ) 1/π (π(β1)+π(β1+β)+π(β1+2β)+ β¦+π(β1+(πβ1)β)) =2 (πππ)β¬(πββ) 1/π (π^(β1)+π^(β1). π^β+π^(β1). π^2β+ β¦+π^(β1).π^(πβ1)β ) =2 (πππ)β¬(πββ) 1/π (π^(β1) (1+β+π^2β+ β¦+π^(πβ1)β )) =2π^(β1) (πππ)β¬(πββ) 1/π (1+β+π^2β+ β¦+π^(πβ1)β ) Let S = 1+β+π^2β+ β¦+π^(πβ1)β It is G.P with common ratio (r) π = π^β/1 = π^β We know Sum of G.P = a((π^π β 1)/(π β 1)) Replacing a by 1 and r by π^π , we get S = 1(((π^β )^π β 1)/(π^π β 1))= (π^πβ β 1)/(π^π β 1) Thus, β«1_(β1)^1βγππ₯ ππ₯γ =2 . π^(β1) (πππ)β¬(πββ) 1/π (1+π^π+π^2β+ β¦+π^(πβ1)β ) = 2/π (πππ)β¬(πββ) 1/π ((π^πβ β 1)/(π^π β 1)) = 2/π (πππ)β¬(πββ) 1/π ((π^πβ β 1)/(β . (π^β β 1)/β)) = 2/π (πππ)β¬(πββ) (π^πβ β 1)/πβ . 1/( (π^β β 1)/β) = 2/π (πππ)β¬(πββ) (π^πβ β 1)/πβ . (πππ)β¬(πβ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 Thus, our equation becomes β«1_(β1)^1βγππ₯ ππ₯γ = 2/π (πππ)β¬(πββ) (π^πβ β 1)/πβ . (πππ)β¬(πββ) 1/( (π^β β 1)/β) = 2/π (πππ)β¬(πββ) (π^πβ β 1)/πβ . 1 = 2/π (πππ)β¬(πββ) (π^(π . 2/π) β 1)/(π (2/π) ) = 2/π (πππ)β¬(πββ) (π^2 β 1)/2 (ππ πππ (πππ)β¬(π‘β0) (π^π‘ β 1)/π‘ =1) (ππ πππ β=2/π) = 2/π . (π^2 β 1)/2 = (π^2 β 1)/π = π^2/π β 1/π =π β π/π