Check sibling questions

Example 26 - Evaluate integral ex dx as the limit of a sum

Example 26 - Chapter 7 Class 12 Integrals - Part 2
Example 26 - Chapter 7 Class 12 Integrals - Part 3
Example 26 - Chapter 7 Class 12 Integrals - Part 4
Example 26 - Chapter 7 Class 12 Integrals - Part 5
Example 26 - Chapter 7 Class 12 Integrals - Part 6
Example 26 - Chapter 7 Class 12 Integrals - Part 7
Example 26 - Chapter 7 Class 12 Integrals - Part 8

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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 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.