Ex 7.8, 2

Transcript

Ex7.8, 2 ﷐0﷮5﷮(𝑥+1) 𝑑𝑥﷯ ﷐0﷮5﷮(𝑥+1) 𝑑𝑥﷯ Putting 𝑎 =0 𝑏 =5 ℎ=﷐𝑏 − 𝑎﷮𝑛﷯ =﷐5 − 0﷮𝑛﷯ =﷐5﷮𝑛﷯ Now, 𝑓﷐𝑥﷯=𝑥+1 𝑓﷐0﷯=0+1=1 𝑓﷐ℎ﷯=ℎ+1 𝑓 ﷐2ℎ﷯=2ℎ+1 …. 𝑓﷐﷐𝑛−1﷯ℎ﷯=﷐𝑛−1﷯ℎ+1 Hence we can write it as ﷐0﷮5﷮(𝑥+1) 𝑑𝑥﷯ =﷐5−0﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯+…+𝑓﷐0+﷐𝑛−1﷯ℎ﷯﷯ =﷐5−0﷯﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯+…+𝑓﷐﷐𝑛−1﷯ℎ﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐1+﷐ℎ+1﷯+﷐2ℎ+1﷯+…+﷐𝑛−1﷯ℎ+1﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐1+1+1+…+1 +ℎ+2ℎ+ …+﷐𝑛−1﷯ℎ﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛 . 1+ℎ ﷐1+2+ …+﷐𝑛−1﷯﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐𝑛+ℎ ﷐𝑛﷐𝑛 − 1﷯﷮2﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1﷮𝑛﷯﷐﷐𝑛﷮𝑛﷯ + ﷐𝑛﷐𝑛 − 1﷯ ℎ﷮2𝑛﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+ ﷐﷐𝑛 − 1﷯ ℎ﷮2﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+ ﷐﷐𝑛 − 1﷯﷮2﷯ . ﷐5﷮𝑛﷯﷯ = 5﷐𝑙𝑖𝑚﷮𝑛→𕔴uc1﷯﷐1+ ﷐5﷮2﷯﷐﷐𝑛 − 1﷮𝑛﷯﷯﷯ = 5﷐1+ ﷐5﷮2﷯﷐1−﷐1﷮𕔴uc1﷯﷯﷯ = 5﷐1+ ﷐5﷮2﷯﷐1−0﷯ ﷯ = 5﷐1+ ﷐5﷮2﷯﷐1−0﷯ ﷯ = 5﷐1+ ﷐5﷮2﷯﷯ = 5 × ﷐7﷮2﷯ = ﷐𝟑𝟓﷮𝟐﷯