Ex 7.8

Chapter 7 Class 12 Integrals
Serial order wise

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Ex 7.8, 2 β«1_0^5βγ(π₯+1) ππ₯γ β«1_0^5βγ(π₯+1) ππ₯γ Putting π =0 π =5 β=(π β π)/π =(5 β 0)/π =5/π π(π₯)=π₯+1 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_0^5βγ(π₯+1) ππ₯γ =(5β0) limβ¬(nββ) 1/π (π(0)+π(0+β)+π(0+2β)+β¦ +π(0+(πβ1)β) =5 limβ¬(nββ) 1/π (π(0)+π(β)+π(2β)β¦β¦+π((πβ1)β) Here, π(π₯)=π₯+1 π(0)=0+1=1 π(β)=β+1 π (2β)=2β+1 β¦. π((πβ1)β)=(πβ1)β+1 Hence, our equation becomes β«1_0^5βγ(π₯+1) ππ₯γ =(5β0) (πππ)β¬(πββ) 1/π (π(0)+π(β)+π(2β)+β¦+π((πβ1)β)) = 5 (πππ)β¬(πββ) 1/π (1+(β+1)+(2β+1)+β¦+(πβ1)β+1) = 5 (πππ)β¬(πββ) 1/π (1+1+1+β¦+1 +β+2β+ β¦+(πβ1)β) = 5 (πππ)β¬(πββ) 1/π (π . 1+β (1+2+ 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 = 5 (πππ)β¬(πββ) 1/π (π+β π(π β 1)/2) = 5 (πππ)β¬(πββ) 1/π (π/π + (π(π β 1) β)/2π) = 5 (πππ)β¬(πββ) (1+ ((π β 1) β)/2) = 5 (πππ)β¬(πββ) (1+ ((π β 1))/2 . 5/π) (ππ πππ β=5/π) = 5 (πππ)β¬(πββ) (1+ 5/2 ((π β 1)/π)) = 5(1+ 5/2 (1β1/β)) = 5(1+ 5/2 (1β0) ) = 5(1+ 5/2 (1β0) ) = 5(1+ 5/2) = 5 Γ 7/2 = ππ/π