Misc 40 - Evaluate 0->1 e2 - 3x dx as a limit of a sum - Definate Integral as a limit of a sum

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Misc 40 Evaluate ﷐0﷮1﷮﷐﷐𝑒﷮2 −3𝑥﷯﷮𝑑𝑥﷯﷯ as a limit of a sum . I1=﷐0﷮1﷮﷐﷐𝑒﷮2 −3𝑥﷯﷮𝑑𝑥﷯﷯ I1=﷐0﷮1﷮﷐﷐𝑒﷮2﷯ . ﷐𝑒﷮−3𝑥﷯﷮𝑑𝑥﷯﷯ I1=﷐𝑒﷮2﷯﷐0﷮1﷮﷐﷐𝑒﷮−3𝑥﷯﷮𝑑𝑥﷯﷯ Let I2=﷐0﷮1﷮﷐﷐𝑒﷮−3𝑥﷯﷮𝑑𝑥﷯﷯ Putting 𝑎 = 0 𝑏 = 1 ℎ=﷐𝑏 − 𝑎﷮𝑛﷯=﷐1 − 0﷮𝑛﷯=﷐1﷮𝑛﷯ We can write it as I2=﷐0﷮1﷮﷐﷐𝑒﷮−3𝑥﷯﷮𝑑𝑥﷯﷯ I2=﷐1−0﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯…𝑓﷐0+﷐𝑛−1﷯ℎ﷯﷯ I2=1 .﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯…𝑓﷐﷐𝑛−1﷯ℎ﷯﷯ 𝑓﷐𝑥﷯=﷐𝑒﷮−3𝑥﷯ 𝑓﷐0﷯=﷐𝑒﷮−3﷐0﷯﷯ =﷐𝑒﷮0﷯=1 𝑓﷐ℎ﷯=﷐𝑒﷮−3ℎ﷯ 𝑓﷐2ℎ﷯=﷐𝑒﷮−3﷐2ℎ﷯﷯=﷐𝑒﷮−6ℎ﷯ …. 𝑓﷐﷐𝑛−1﷯ℎ﷯=﷐𝑒﷮−3﷐𝑛 − 1﷯ℎ﷯ ∴ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐1+﷐𝑒﷮−3ℎ﷯+﷐𝑒﷮−6ℎ﷯…+ ﷐𝑒﷮−3﷐𝑛 − 1﷯ℎ﷯﷯ Let 𝑆=1+﷐𝑒﷮−3ℎ﷯+﷐𝑒﷮−6ℎ﷯…+ ﷐𝑒﷮−3﷐𝑛 − 1﷯ℎ﷯ If is G.P with common ratio (r) 𝑟 = ﷐﷐𝑒﷮−6ℎ﷯﷮﷐𝑒﷮−3ℎ﷯﷯ = ﷐𝑒﷮−3ℎ﷯ We know, Sum of G.P, S = 𝑎﷐﷐﷐𝑟﷮𝑛﷯ − 1﷮𝑟 − 1﷯﷯ Replacing a by 1 an r by ﷐𝑒﷮−3ℎ﷯, we get S =1﷐﷐﷐﷐﷐𝑒﷮−3ℎ﷯﷯﷮𝑛﷯ − 1﷮﷐𝑒﷮−3ℎ﷯ − 1﷯﷯=﷐﷐𝑒﷮−3𝑛ℎ﷯ − 1﷮﷐𝑒﷮−3ℎ﷯ − 1﷯ Thus, ∴ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐1+﷐𝑒﷮−3ℎ﷯+﷐𝑒﷮−6ℎ﷯…+ ﷐𝑒﷮−3﷐𝑛 − 1﷯ℎ﷯﷯ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐﷐﷐𝑒﷮−3𝑛ℎ﷯ − 1﷮﷐𝑒﷮−3ℎ﷯ − 1﷯﷯ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐﷐﷐𝑒﷮−3𝑛ℎ﷯ − 1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯ × −3ℎ ﷯﷯ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮−3𝑛ℎ﷯ ﷐﷐﷐𝑒﷮−3𝑛ℎ﷯ − 1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯﷯﷯ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮−3𝑛ℎ﷯ ﷐﷐𝑒﷮−3𝑛ℎ﷯−1﷯ .﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐﷐1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯﷯﷯ Taking, ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐﷐1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯﷯﷯ As 𝑛→𕔴 ⇒﷐1﷮ℎ﷯→𕔴 ⇒ℎ→0 ⇒−3ℎ→0 ∴ ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐﷐1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯﷯﷯ =﷐𝑙𝑖𝑚﷮−3ℎ→0﷯﷐﷐1﷮﷐﷐𝑒﷮−3ℎ﷯ − 1﷮−3ℎ﷯﷯﷯ =﷐1﷮1﷯ =1 Thus, ∴ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮−3𝑛ℎ﷯ ﷐﷐𝑒﷮−3𝑛ℎ﷯−1﷯ . 1 I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐﷐𝑒﷮−3𝑛﷐﷐1﷮𝑛﷯﷯﷯ − 1﷮−3𝑛﷐﷐1﷮𝑛﷯﷯﷯﷯ I2=﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐﷐𝑒﷮−3﷯ − 1﷮−3﷯﷯ I2= ﷐﷐𝑒﷮−3﷯ − 1﷮−3﷯ I2= ﷐1 − ﷐𝑒﷮−3﷯﷮3﷯ I2=﷐1﷮3﷯﷐1− ﷐1﷮﷐𝑒﷮−3﷯﷯﷯ I2=﷐1﷮3﷯﷐﷐﷐𝑒﷮3﷯ − 1﷮﷐𝑒﷮3﷯﷯﷯ Putting the value of I2 in I1 i.e., eq (1) we get I1=﷐𝑒﷮2﷯×﷐1﷮3﷯﷐﷐﷐𝑒﷮3﷯ − 1﷮﷐𝑒﷮3﷯﷯﷯ I1=﷐1﷮3﷯﷐﷐﷐𝑒﷮3﷯ − 1﷮𝑒﷯﷯ ∴ 𝐈𝟏=﷐𝟏﷮𝟑﷯﷐﷐𝒆﷮𝟐﷯− ﷐𝟏﷮𝒆﷯﷯

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