Check sibling questions

Example 25 - Find integral (x2 + 1)dx as limit of a sum - Examples

Example 25 - Chapter 7 Class 12 Integrals - Part 2
Example 25 - Chapter 7 Class 12 Integrals - Part 3 Example 25 - Chapter 7 Class 12 Integrals - Part 4 Example 25 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Example 25 Find ∫_0^2β–’(π‘₯^2+1) 𝑑π‘₯ as the limit of a sum . ∫_0^2β–’(π‘₯^2+1) 𝑑π‘₯ Putting π‘Ž = 0 𝑏 = 2 β„Ž = (𝑏 βˆ’ π‘Ž)/𝑛 = (2 βˆ’ 0)/𝑛 = 2/𝑛 𝑓(π‘₯)=π‘₯^2+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 ∫_0^2β–’(π‘₯^2+1) 𝑑π‘₯ =(2βˆ’0) lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(0+β„Ž)+𝑓(0+2β„Ž)+… +𝑓(0+(π‘›βˆ’1)β„Ž) =2 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓((π‘›βˆ’1)β„Ž) Here, 𝑓(π‘₯)=π‘₯^2+1 𝑓(0)=0^2+1=0+1=1 𝑓(β„Ž)=β„Ž^2+1=(2/𝑛)^2+1=4/𝑛^2 +1 𝑓(2β„Ž)=(2β„Ž)^2+1=γ€–4β„Žγ€—^2+1=4(2/𝑛)^2+1=16/𝑛^2 +1 ….. 𝑓(π‘›βˆ’1)β„Ž=((π‘›βˆ’1)β„Ž)^2+1=γ€–(π‘›βˆ’1)^2 (2/𝑛)γ€—^2+1 =(π‘›βˆ’1)^2 Γ— 4/𝑛^2 +1 Hence, our equation becomes = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑓(0)+𝑓(β„Ž)+𝑓(2β„Ž)……+𝑓(π‘›βˆ’1)β„Ž) = 2 lim┬(nβ†’βˆž) 1/𝑛 (1+(4/𝑛^2 +1)+(16/𝑛^2 +1" " )+ ……+((4(𝑛 βˆ’ 1)^2)/𝑛^2 +1)) = 2 lim┬(nβ†’βˆž) 1/𝑛 ((1 + 1 + 1…𝑛 π‘‘π‘–π‘šπ‘’π‘ )+0+ 4/𝑛^2 +16/𝑛^2 + …(4(𝑛 βˆ’ 1)^2)/𝑛^2 ) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛 +0+ 4/𝑛^2 +16/𝑛^2 + ……(4(𝑛 βˆ’ 1)^2)/𝑛^2 ) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛+ 4/𝑛^2 (1+4+ ……+(𝑛 βˆ’ 1)^2 ) ) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛+ 4/𝑛^2 (1^2+2^2+ ………+(𝑛 βˆ’ 1)^2 ) ) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛+ 4/𝑛^2 ((𝑛 βˆ’ 1) 𝑛(2𝑛 βˆ’ 1))/6) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛+ 4/𝑛 ((𝑛 βˆ’ 1) (2𝑛 βˆ’ 1))/6) = 2 lim┬(nβ†’βˆž) 1/𝑛 (𝑛+ 2/3𝑛 (π‘›βˆ’1) (2π‘›βˆ’1)) = 2 lim┬(nβ†’βˆž) (𝑛/𝑛 + 2/(3𝑛^2 ) (π‘›βˆ’1) (2π‘›βˆ’1)) We know that 1^2+2^2+ ……+𝑛^2= (𝑛(𝑛 + 1) (2𝑛 +1))/6 1^2+2^2+ ……+(π‘›βˆ’1)^2= ((𝑛 βˆ’ 1)(𝑛 βˆ’ 1+ 1) (2(𝑛 βˆ’1)+1))/6 = ((𝑛 βˆ’ 1) 𝑛(2𝑛 βˆ’ 1))/6 = 2 lim┬(nβ†’βˆž) (𝑛/𝑛 + 2/3 ((𝑛 βˆ’ 1))/𝑛 ((2𝑛 βˆ’ 1))/𝑛) = 2 lim┬(nβ†’βˆž) (1+ 2/3 (1βˆ’ 1/𝑛) (2βˆ’ 1/𝑛)) = 2 (1+ 2/3 (1βˆ’0) (2βˆ’0)) = 2 (1+ 2/3 Γ—2) = 2 (1+ 4/3) = 2 Γ— 7/3 = πŸπŸ’/πŸ‘ (lim┬(nβ†’βˆž) 1/𝑛=0" " )

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.