Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Transcript

Example 26 Evaluate ∫_0^2▒𝑒^π‘₯ 𝑑π‘₯ as the limit of a sum . ∫_0^2▒𝑒^π‘₯ 𝑑π‘₯ Putting π‘Ž = 0 𝑏 = 2 β„Ž = (𝑏 βˆ’ π‘Ž)/𝑛 = (2 βˆ’ 0)/𝑛 = 2/𝑛 𝑓(π‘₯)=𝑒^π‘₯ 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 ∫_0^2▒𝑒^π‘₯ 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‘₯)=𝑒^π‘₯ 𝑓(0)=𝑒^0=1 𝑓(0+β„Ž)=𝑒^(0 + β„Ž)=𝑒^β„Ž 𝑓(0+2β„Ž)=𝑒^(0 + 2β„Ž)=𝑒^2β„Ž 𝑓(0+(π‘›βˆ’1)β„Ž)=𝑒^(0 + (π‘›βˆ’1)β„Ž)=𝑒^(π‘›βˆ’1)β„Ž Hence, our equation becomes ∴ ∫_0^2▒𝑒^π‘₯ 𝑑π‘₯ =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+𝑒^β„Ž+𝑒^2β„Ž+ ……+𝑒^((𝑛 βˆ’ 1) β„Ž) ) Let S = 𝑒^β„Ž+𝑒^2β„Ž+ ……+𝑒^((𝑛 βˆ’ 1) β„Ž) It is a G.P. with common ratio (r) r = 𝑒^2β„Ž/𝑒^β„Ž = 𝑒^(2β„Ž βˆ’ β„Ž) = 𝑒^β„Ž Sum of G.P. S = (π‘Ž (1 βˆ’ π‘Ÿ^𝑛 ))/(1 βˆ’ π‘Ÿ) = (𝑒^β„Ž (1 βˆ’ (𝑒^β„Ž )^𝑛 ))/(1 βˆ’ 𝑒^β„Ž ) = (𝑒^β„Ž (1 βˆ’ 𝑒^π‘›β„Ž ))/(1 βˆ’ 𝑒^β„Ž ) = (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 (1+𝑒^β„Ž+𝑒^2β„Ž+ ……+𝑒^((𝑛 βˆ’ 1) β„Ž) ) Let S = 𝑒^β„Ž+𝑒^2β„Ž+ ……+𝑒^((𝑛 βˆ’ 1) β„Ž) It is a G.P. with common ratio (r) r = 𝑒^2β„Ž/𝑒^β„Ž = 𝑒^(2β„Ž βˆ’ β„Ž) = 𝑒^β„Ž Sum of G.P. S = (π‘Ž (1 βˆ’ π‘Ÿ^𝑛 ))/(1 βˆ’ π‘Ÿ) = (𝑒^β„Ž (1 βˆ’ (𝑒^β„Ž )^𝑛 ))/(1 βˆ’ 𝑒^β„Ž ) = (𝑒^β„Ž (1 βˆ’ 𝑒^π‘›β„Ž ))/(1 βˆ’ 𝑒^β„Ž ) Thus ∴ ∫_0^2▒𝑒^π‘₯ 𝑑π‘₯ = 2 .lim┬(nβ†’βˆž) 1/𝑛 (1+𝑒^β„Ž+𝑒^2β„Ž+ ……+𝑒^((𝑛 βˆ’ 1) β„Ž) ) Putting the value of S, we get = 2 .lim┬(nβ†’βˆž) 1/𝑛 (1+ (𝑒^β„Ž (1 βˆ’ 𝑒^π‘›β„Ž ))/(1 βˆ’ 𝑒^β„Ž )) = 2 (lim┬(nβ†’βˆž) 1/𝑛 +lim┬(nβ†’βˆž) 1/𝑛 (𝑒^β„Ž ((1 βˆ’ 𝑒^π‘›β„Ž)/(1 βˆ’ 𝑒^β„Ž )))) "Taking βˆ’1 common from" "numerator and denominator " = 2 (1/∞ +lim┬(nβ†’βˆž) (𝑒^β„Ž/𝑛 . ( 𝑒^π‘›β„Ž βˆ’ 1)/(𝑒^β„Ž βˆ’ 1))) "Multiplying and dividing denominator by h" = 2 (0+lim┬(nβ†’βˆž) (𝑒^β„Ž/𝑛 . ( 𝑒^π‘›β„Ž βˆ’ 1)/(β„Ž . ( 𝑒^β„Ž βˆ’ 1)/β„Ž))) = 2 . lim┬(nβ†’βˆž) 𝑒^β„Ž/𝑛 (( 𝑒^π‘›β„Ž βˆ’ 1)/β„Ž)(( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž)) = 2 . lim┬(nβ†’βˆž) 𝑒^β„Ž/𝑛 (( 𝑒^π‘›β„Ž βˆ’ 1)/β„Ž) . lim┬(nβ†’βˆž) (( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž)) Solving (π₯𝐒𝐦)┬(π§β†’βˆž) ( 𝟏)/(( 𝒆^𝒉 βˆ’ 𝟏)/𝒉) As nβ†’βˆž β‡’ 2/β„Ž β†’βˆž β‡’ β„Ž β†’0 ∴ lim┬(nβ†’βˆž) ( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž) = lim┬(hβ†’0) ( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž) = 1/1 = 1 (π‘ˆπ‘ π‘–π‘›π‘” lim┬(t β†’ 0) ( 𝑒^𝑑 βˆ’ 1)/𝑑=1) Thus, our equation becomes ∴ ∫_0^2▒𝑒^π‘₯ 𝑑π‘₯ = 2 . lim┬(nβ†’βˆž) 𝑒^β„Ž/𝑛 (( 𝑒^π‘›β„Ž βˆ’ 1)/β„Ž) . lim┬(nβ†’βˆž) (( 1)/(( 𝑒^β„Ž βˆ’ 1)/β„Ž)) = 2 . lim┬(nβ†’βˆž) 𝑒^β„Ž/𝑛 (( 𝑒^π‘›β„Ž βˆ’ 1)/β„Ž) . 1 = 2 . lim┬(nβ†’βˆž) 𝑒^( 2/𝑛)/𝑛 (( 𝑒^(𝑛 . 2/𝑛) βˆ’ 1)/(2/𝑛)) = 2 . lim┬(nβ†’βˆž) γ€– 𝑒〗^( 2/𝑛) (( 𝑒^2 βˆ’ 1)/2) = 2 (𝑒^( 2/∞) ((𝑒^2 βˆ’ 1))/2) (π‘ˆπ‘ π‘–π‘›π‘” β„Ž= 2/𝑛) = 2 . 𝑒^0 ((𝑒^2 βˆ’ 1))/2 = 2/2 . 1 (𝑒^2βˆ’1) = 1 . 1 (𝑒^2βˆ’1) = 𝒆^πŸβˆ’πŸ

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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.