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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.8, 2 ∫1_0^5β–’γ€–(π‘₯+1) 𝑑π‘₯γ€— ∫1_0^5β–’γ€–(π‘₯+1) 𝑑π‘₯γ€— Putting π‘Ž =0 𝑏 =5 β„Ž=(𝑏 βˆ’ π‘Ž)/𝑛 =(5 βˆ’ 0)/𝑛 =5/𝑛 𝑓(π‘₯)=π‘₯+1 We know that ∫1_π‘Ž^𝑏▒〖π‘₯ 𝑑π‘₯γ€— =(π‘βˆ’π‘Ž) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(π‘Ž)+𝑓(π‘Ž+β„Ž)+𝑓(π‘Ž+2β„Ž)…+𝑓(π‘Ž+(π‘›βˆ’1)β„Ž)) Hence we can write ∫1_0^5β–’γ€–(π‘₯+1) 𝑑π‘₯γ€— =(5βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =5 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=π‘₯+1 𝑓(0)=0+1=1 𝑓(β„Ž)=β„Ž+1 𝑓 (2β„Ž)=2β„Ž+1 …. 𝑓((π‘›βˆ’1)β„Ž)=(π‘›βˆ’1)β„Ž+1 Hence, our equation becomes ∫1_0^5β–’γ€–(π‘₯+1) 𝑑π‘₯γ€— =(5βˆ’0) (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)+…+𝑓((π‘›βˆ’1)β„Ž)) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+(β„Ž+1)+(2β„Ž+1)+…+(π‘›βˆ’1)β„Ž+1) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (1+1+1+…+1 +β„Ž+2β„Ž+ …+(π‘›βˆ’1)β„Ž) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑛 . 1+β„Ž (1+2+ …+(π‘›βˆ’1))) We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+π‘›βˆ’1= ((𝑛 βˆ’ 1) (𝑛 βˆ’ 1 + 1))/2 = (𝑛 (𝑛 βˆ’ 1) )/2 = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑛+β„Ž 𝑛(𝑛 βˆ’ 1)/2) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) 1/𝑛 (𝑛/𝑛 + (𝑛(𝑛 βˆ’ 1) β„Ž)/2𝑛) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+ ((𝑛 βˆ’ 1) β„Ž)/2) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+ ((𝑛 βˆ’ 1))/2 . 5/𝑛) (π‘ˆπ‘ π‘–π‘›π‘” β„Ž=5/𝑛) = 5 (π‘™π‘–π‘š)┬(π‘›β†’βˆž) (1+ 5/2 ((𝑛 βˆ’ 1)/𝑛)) = 5(1+ 5/2 (1βˆ’1/∞)) = 5(1+ 5/2 (1βˆ’0) ) = 5(1+ 5/2 (1βˆ’0) ) = 5(1+ 5/2) = 5 Γ— 7/2 = πŸ‘πŸ“/𝟐

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.