Example 25 - Chapter 7 Class 12 Integrals

Transcript

Example 25 Find ﷐0﷮2﷮﷐﷐𝑥﷮2﷯+1﷯﷯ 𝑑𝑥 as the limit of a sum . ﷐0﷮2﷮﷐﷐𝑥﷮2﷯+1﷯﷯ 𝑑𝑥 Putting 𝑎 = 0 𝑏 = 2 ℎ = ﷐𝑏 − 𝑎﷮𝑛﷯ = ﷐2 − 0﷮𝑛﷯ = ﷐2﷮𝑛﷯ 𝑓﷐𝑥﷯=﷐𝑥﷮2﷯+1 Hence we can write ﷐0﷮2﷮﷐﷐𝑥﷮2﷯+1﷯﷯ 𝑑𝑥 =﷐2−0﷯﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐0+ℎ﷯+𝑓﷐0+2ℎ﷯+… +𝑓0+﷐𝑛−1﷯ℎ﷯ =2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯……+𝑓﷐𝑛−1﷯ℎ﷯ 𝑓﷐𝑥﷯=﷐𝑥﷮2﷯+1 𝑓﷐0﷯=﷐0﷮2﷯+1=0+1=1 𝑓﷐ℎ﷯=﷐ℎ﷮2﷯+1=﷐﷐﷐2﷮𝑛﷯﷯﷮2﷯+1=﷐4﷮﷐𝑛﷮2﷯﷯ +1 𝑓﷐2ℎ﷯=﷐﷐2ℎ﷯﷮2﷯+1=﷐4ℎ﷮2﷯+1=﷐4﷐﷐2﷮𝑛﷯﷯﷮2﷯+1=﷐16﷮﷐𝑛﷮2﷯﷯ +1 ….. 𝑓﷐𝑛−1﷯ℎ=﷐﷐﷐𝑛−1﷯ℎ﷯﷮2﷯+1=﷐﷐﷐𝑛−1﷯﷮2﷯﷐﷐2﷮𝑛﷯﷯﷮2﷯+1 =﷐﷐𝑛−1﷯﷮2﷯ × ﷐4﷮﷐𝑛﷮2﷯﷯ +1 Hence, our equation becomes = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑓﷐0﷯+𝑓﷐ℎ﷯+𝑓﷐2ℎ﷯……+𝑓﷐𝑛−1﷯ℎ﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐1+﷐﷐4﷮﷐𝑛﷮2﷯﷯ +1﷯+﷐﷐16﷮﷐𝑛﷮2﷯﷯ +1 ﷯+ ……+﷐﷐4﷐﷐𝑛 − 1﷯﷮2﷯﷮﷐𝑛﷮2﷯﷯+1﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐﷐1 + 1 + 1…𝑛 𝑡𝑖𝑚𝑒𝑠﷯+0+ ﷐4﷮﷐𝑛﷮2﷯﷯ +﷐16﷮﷐𝑛﷮2﷯﷯ + …﷐4﷐﷐𝑛 − 1﷯﷮2﷯﷮﷐𝑛﷮2﷯﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛 +0+ ﷐4﷮﷐𝑛﷮2﷯﷯ +﷐16﷮﷐𝑛﷮2﷯﷯+ ……﷐4﷐﷐𝑛 − 1﷯﷮2﷯﷮﷐𝑛﷮2﷯﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛+ ﷐4﷮﷐𝑛﷮2﷯﷯ ﷐1+4+ ……+﷐﷐𝑛 − 1﷯﷮2﷯﷯ ﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛+ ﷐4﷮﷐𝑛﷮2﷯﷯ ﷐﷐1﷮2﷯+﷐2﷮2﷯+ ………+﷐﷐𝑛 − 1﷯﷮2﷯﷯ ﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛+ ﷐4﷮﷐𝑛﷮2﷯﷯ ﷐﷐𝑛 − 1﷯ 𝑛﷐2𝑛 − 1﷯﷮6﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛+ ﷐4﷮𝑛﷯ ﷐﷐𝑛 − 1﷯ ﷐2𝑛 − 1﷯﷮6﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1﷮𝑛﷯﷐𝑛+ ﷐2﷮3𝑛﷯ ﷐𝑛−1﷯ ﷐2𝑛−1﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐﷐𝑛﷮𝑛﷯ + ﷐2﷮3﷐𝑛﷮2﷯﷯ ﷐𝑛−1﷯ ﷐2𝑛−1﷯﷯ = 2 ﷐lim﷮n→𕔴uc1﷯ ﷐1+ ﷐2﷮3﷯ ﷐1− ﷐1﷮𝑛﷯﷯ ﷐2− ﷐1﷮𝑛﷯﷯﷯ = 2 ﷐1+ ﷐2﷮3﷯ ﷐1−0﷯ ﷐2−0﷯﷯ = 2 ﷐1+ ﷐2﷮3﷯ ×2﷯ = 2 ﷐1+ ﷐4﷮3﷯﷯ = 2 × ﷐7﷮3﷯ = ﷐𝟏𝟒﷮𝟑﷯