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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 1 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 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.