Area as a sum

Chapter 7 Class 12 Integrals
Serial order wise

Transcript

Question 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

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.