Examples

Chapter 7 Class 12 Integrals
Serial order wise

Transcript

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" " )

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.