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Question 1 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 We know that ∫1_𝑎^𝑏▒〖𝑥 𝑑𝑥〗 =(𝑏−𝑎) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(𝑎)+𝑓(𝑎+ℎ)+𝑓(𝑎+2ℎ)…+𝑓(𝑎+(𝑛−1)ℎ)) Hence we can write ∫_0^2▒(𝑥^2+1) 𝑑𝑥 =(2−0) lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(0+ℎ)+𝑓(0+2ℎ)+… +𝑓(0+(𝑛−1)ℎ) =2 lim┬(n→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓((𝑛−1)ℎ) Here, 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→∞) 1/𝑛 (𝑓(0)+𝑓(ℎ)+𝑓(2ℎ)……+𝑓(𝑛−1)ℎ) = 2 lim┬(n→∞) 1/𝑛 (1+(4/𝑛^2 +1)+(16/𝑛^2 +1" " )+ ……+((4(𝑛 − 1)^2)/𝑛^2 +1)) = 2 lim┬(n→∞) 1/𝑛 ((1 + 1 + 1…𝑛 𝑡𝑖𝑚𝑒𝑠)+0+ 4/𝑛^2 +16/𝑛^2 + …(4(𝑛 − 1)^2)/𝑛^2 ) = 2 lim┬(n→∞) 1/𝑛 (𝑛 +0+ 4/𝑛^2 +16/𝑛^2 + ……(4(𝑛 − 1)^2)/𝑛^2 ) = 2 lim┬(n→∞) 1/𝑛 (𝑛+ 4/𝑛^2 (1+4+ ……+(𝑛 − 1)^2 ) ) = 2 lim┬(n→∞) 1/𝑛 (𝑛+ 4/𝑛^2 (1^2+2^2+ ………+(𝑛 − 1)^2 ) ) = 2 lim┬(n→∞) 1/𝑛 (𝑛+ 4/𝑛^2 ((𝑛 − 1) 𝑛(2𝑛 − 1))/6) = 2 lim┬(n→∞) 1/𝑛 (𝑛+ 4/𝑛 ((𝑛 − 1) (2𝑛 − 1))/6) = 2 lim┬(n→∞) 1/𝑛 (𝑛+ 2/3𝑛 (𝑛−1) (2𝑛−1)) = 2 lim┬(n→∞) (𝑛/𝑛 + 2/(3𝑛^2 ) (𝑛−1) (2𝑛−1)) We know that 1^2+2^2+ ……+𝑛^2= (𝑛(𝑛 + 1) (2𝑛 +1))/6 1^2+2^2+ ……+(𝑛−1)^2= ((𝑛 − 1)(𝑛 − 1+ 1) (2(𝑛 −1)+1))/6 = ((𝑛 − 1) 𝑛(2𝑛 − 1))/6 = 2 lim┬(n→∞) (𝑛/𝑛 + 2/3 ((𝑛 − 1))/𝑛 ((2𝑛 − 1))/𝑛) = 2 lim┬(n→∞) (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 (lim┬(n→∞) 1/𝑛=0" " )

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo