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Ex 7.8, 2 - Integrate (x + 1) dx from 0 to 5 by limit as a sum

Ex 7.8, 2 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.8, 2 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.8, 2 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.8, 2 - Chapter 7 Class 12 Integrals - Part 5

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.