


Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex7.8, 2 05(𝑥+1) 𝑑𝑥 05(𝑥+1) 𝑑𝑥 Putting 𝑎 =0 𝑏 =5 ℎ=𝑏 − 𝑎𝑛 =5 − 0𝑛 =5𝑛 Now, 𝑓𝑥=𝑥+1 𝑓0=0+1=1 𝑓ℎ=ℎ+1 𝑓 2ℎ=2ℎ+1 …. 𝑓𝑛−1ℎ=𝑛−1ℎ+1 Hence we can write it as 05(𝑥+1) 𝑑𝑥 =5−0𝑙𝑖𝑚𝑛→uc11𝑛𝑓0+𝑓0+ℎ+𝑓0+2ℎ+…+𝑓0+𝑛−1ℎ =5−0𝑙𝑖𝑚𝑛→uc11𝑛𝑓0+𝑓ℎ+𝑓2ℎ+…+𝑓𝑛−1ℎ = 5𝑙𝑖𝑚𝑛→uc11𝑛1+ℎ+1+2ℎ+1+…+𝑛−1ℎ+1 = 5𝑙𝑖𝑚𝑛→uc11𝑛1+1+1+…+1 +ℎ+2ℎ+ …+𝑛−1ℎ = 5𝑙𝑖𝑚𝑛→uc11𝑛𝑛 . 1+ℎ 1+2+ …+𝑛−1 = 5𝑙𝑖𝑚𝑛→uc11𝑛𝑛+ℎ 𝑛𝑛 − 12 = 5𝑙𝑖𝑚𝑛→uc11𝑛𝑛𝑛 + 𝑛𝑛 − 1 ℎ2𝑛 = 5𝑙𝑖𝑚𝑛→uc11+ 𝑛 − 1 ℎ2 = 5𝑙𝑖𝑚𝑛→uc11+ 𝑛 − 12 . 5𝑛 = 5𝑙𝑖𝑚𝑛→uc11+ 52𝑛 − 1𝑛 = 51+ 521−1uc1 = 51+ 521−0 = 51+ 521−0 = 51+ 52 = 5 × 72 = 𝟑𝟓𝟐
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