Ex 7.8, 6 -  Integrate (x + e2x) dx from 0 to 4 by limit as a sum  - Ex 7.8

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex7.8, 6 ﷐0﷮4﷮﷐𝑥+𝑒2𝑥﷯𝑑𝑥﷯ Let I1 = ﷐0﷮4﷮﷐𝑥+𝑒2𝑥﷯𝑑𝑥﷯ I1 = ﷐0﷮4﷮𝑥 . 𝑑𝑥﷯+﷐0﷮4﷮ 𝑒2𝑥 . 𝑑𝑥﷯ Solving I2 I2 = ﷐0﷮4﷮𝑥 . 𝑑𝑥﷯ Putting 𝑎 =0 𝑏 =4 ℎ=﷐𝑏 − 𝑎﷮𝑛﷯ = ﷐4 − 0﷮𝑛﷯ =﷐4﷮𝑛﷯ & 𝑓﷐𝑥﷯=𝑥 We can write it as ﷐0﷮4﷮𝑥 . 𝑑𝑥﷯ =﷐4 − 0﷯ ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯+ …+𝑓﷐0+﷐𝑛−1﷯ℎ﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯+ …+𝑓﷐﷐𝑛−1﷯ℎ﷯﷯ 𝑓﷐𝑥﷯=𝑥 𝑓﷐0﷯=0 𝑓﷐0+ℎ﷯=0+ℎ=ℎ 𝑓﷐0+2ℎ﷯=0+2ℎ=2ℎ … 𝑓﷐0+﷐𝑛−1﷯ℎ﷯=0+﷐𝑛−1﷯ℎ=﷐𝑛−1﷯ℎ Thus, ﷐0﷮4﷮𝑥 . 𝑑𝑥﷯ =﷐4 − 0﷯ ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯+ …+𝑓﷐0+﷐𝑛−1﷯ℎ﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮𝑛﷯ ﷐0+ℎ+2ℎ+ …+﷐𝑛−1﷯ℎ﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮𝑛﷯ ﷐ℎ﷐1+2+3+ …+﷐𝑛−1﷯﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮𝑛﷯ ﷐﷐ℎ 𝑛 ﷐𝑛 − 1﷯﷮2﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐﷐𝑛 − 1﷯ ℎ﷮2﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑛 − 1﷮2﷯ × ﷐4﷮𝑛﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑛 − 1﷮𝑛﷯ × ﷐4﷮2﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐1 − ﷐1﷮𝑛﷯﷯2﷯ =4 . 2 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1 − ﷐1﷮𝑛﷯﷯ =8﷐1 − ﷐1﷮𕔴uc1﷯﷯ =8﷐1−0﷯ =8 Solving I3 I3 = ﷐0﷮4﷮﷐𝑒﷮2𝑥﷯ . 𝑑𝑥﷯ Putting 𝑎 =0 𝑏 =4 ℎ=﷐𝑏 − 𝑎﷮𝑛﷯ = ﷐4 − 0﷮𝑛﷯ =﷐4﷮𝑛﷯ & 𝑓﷐𝑥﷯=﷐𝑒﷮2𝑥﷯ We can write it as ﷐0﷮4﷮﷐𝑒﷮2𝑥﷯ . 𝑑𝑥﷯ =﷐4 − 0﷯ ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯+ …+𝑓﷐0+﷐𝑛−1﷯ℎ﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯+ …+𝑓﷐﷐𝑛−1﷯ℎ﷯﷯ 𝑓﷐𝑥﷯=﷐𝑒﷮2𝑥﷯ 𝑓﷐0﷯=﷐𝑒﷮2 . 0﷯= ﷐𝑒﷮0﷯=1 𝑓﷐0+ℎ﷯=﷐𝑒﷮2﷐0 + ℎ﷯﷯= ﷐𝑒﷮2ℎ﷯ 𝑓﷐0+2ℎ﷯=﷐𝑒﷮2﷐0 + 2ℎ﷯﷯= ﷐𝑒﷮2﷐2ℎ﷯﷯ …. 𝑓﷐0+﷐𝑛−1﷯ℎ﷯=﷐𝑒﷮2﷐0 + ﷐𝑛−1﷯ℎ﷯﷯= ﷐𝑒﷮2﷐﷐𝑛−1﷯ℎ﷯﷯ ∴ ﷐0﷮4﷮﷐𝑒﷮2𝑥﷯ . 𝑑𝑥﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯+ …+𝑓﷐﷐𝑛−1﷯ℎ﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯ ﷐1+﷐𝑒﷮2ℎ﷯+﷐𝑒﷮2﷐2ℎ﷯﷯+ …+﷐𝑒﷮2﷐﷐𝑛−1﷯ℎ﷯﷯﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1+﷐𝑒﷮2ℎ﷯+﷐𝑒﷮4ℎ﷯+ …+﷐𝑒﷮2ℎ﷐𝑛−1﷯﷯﷯ Let S = 1+﷐𝑒﷮2ℎ﷯+﷐𝑒﷮4ℎ﷯+ …+﷐𝑒﷮2ℎ﷐𝑛−1﷯﷯ It is G.P with common ratio (r) , 𝑟 = ﷐﷐𝑒﷮4ℎ﷯﷮﷐𝑒﷮2ℎ﷯﷯ = ﷐𝑒﷮4ℎ −2ℎ﷯ = ﷐𝑒﷮2ℎ﷯ We know, Sum of G.P = a﷐﷐﷐𝑟﷮𝑛﷯ − 1﷮𝑟 − 1﷯﷯ Replacing a by 1 and r by ﷐𝑒﷮2ℎ﷯ , we get S = 1﷐﷐﷐﷐﷐𝑒﷮2ℎ﷯﷯﷮𝑛﷯ − 1﷮﷐𝑒﷮2ℎ﷯ − 1﷯﷯ = ﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮﷐𝑒﷮2ℎ﷯ − 1﷯ Thus, ﷐0﷮4﷮﷐𝑒﷮2𝑥﷯. 𝑑𝑥﷯ =4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1+﷐𝑒﷮2ℎ﷯+﷐𝑒﷮4ℎ﷯+ …+﷐𝑒﷮2ℎ﷐𝑛−1﷯﷯﷯ = 4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮𝑛﷯ ﷐﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮﷐𝑒﷮2ℎ﷯ − 1﷯﷯ = 4 ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮𝑛﷯ ﷐﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯﷯﷐1﷮2ℎ﷯ = 4 ﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮2𝑛 ℎ﷯﷯﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯﷯ Taking ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯ As, 𝑛→𕔴 ⇒ ﷐4﷮𝑛﷯→0 ⇒ ℎ→0 ⇒ 2ℎ→0 ∴ ﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯ = ﷐𝑙𝑖𝑚﷮2ℎ→0﷯ ﷐1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯ = ﷐1﷮1﷯ = 1 Thus, ﷐0﷮4﷮﷐𝑒﷮2𝑥﷯. 𝑑𝑥﷯ = 4 ﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮2𝑛 ℎ﷯﷯﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐1﷮﷐﷐𝑒﷮2ℎ﷯ − 1﷮2ℎ﷯﷯﷯ = 4 ﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑒﷮2𝑛ℎ﷯ − 1﷮2𝑛 ℎ﷯﷯.1 = 4 ﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑒﷮2𝑛﷐﷐4﷮𝑛﷯﷯﷯ − 1﷮2﷐𝑛﷯ ﷐﷐4﷮𝑛﷯﷯﷯﷯.1 = 4 ﷐﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯ ﷐﷐𝑒﷮8﷯ − 1﷮8﷯﷯ = 4 ﷐﷐﷐𝑒﷮8﷯ − 1﷮8﷯﷯ = ﷐﷐𝑒﷮8﷯ − 1﷮2﷯ Putting the values of I2 and I3 in I1 ∴ I1 = ﷐0﷮4﷮𝑥 . 𝑑𝑥﷯+﷐0﷮4﷮ 𝑒2𝑥 . 𝑑𝑥﷯ = 8 + ﷐﷐𝑒﷮8﷯ − 1﷮2﷯ = ﷐16 + ﷐𝑒﷮8﷯ − 1﷮2﷯ = ﷐𝟏𝟓 + ﷐𝒆﷮𝟖﷯﷮𝟐﷯

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