Question 1 - Find integral (x2 + 1)dx as limit of a sum

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" " )

Question 1 - Chapter 7 Class 12 Integrals (Important Question)

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Davneet Singh

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