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Ex 7.8, 3 Class 12 Math - Integrate x^2 from 2 to 3 by limit as a sum

Ex 7.8, 3 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.8, 3 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.8, 3 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.8, 3 - Chapter 7 Class 12 Integrals - Part 5
Ex 7.8, 3 - Chapter 7 Class 12 Integrals - Part 6

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Transcript

Ex 7.8, 3 ∫1_2^3▒〖𝑥2 𝑑𝑥〗 ∫1_2^3▒〖𝑥2 𝑑𝑥〗 Putting 𝑎 =2 𝑏 =3 ℎ=(𝑏 − 𝑎)/𝑛 =(3 − 2)/𝑛 =1/𝑛 𝑓(𝑥)=𝑥^2 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 ∫1_2^3▒〖𝑥2 𝑑𝑥〗 =(3−2) (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(2)+𝑓(2+ℎ)+𝑓(2+2ℎ)+ …+𝑓(2+(𝑛−1)ℎ)) =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(2)+𝑓(2+ℎ)+𝑓(2+2ℎ)+ …+𝑓(2+(𝑛−1)ℎ)) Here, 𝑓(𝑥)=𝑥^2 𝑓(2)=(2)^2=4 𝑓(2+ℎ)=(2+ℎ)^2 𝑓 (2+2ℎ)=(2+2ℎ)^2 … 𝑓(2+(𝑛−1)ℎ)=(2+(𝑛−1)ℎ)^2 Hence, our equation becomes ∫1_2^3▒〖𝑥2 𝑑𝑥〗 " " =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑓(2)+𝑓(2+ℎ)+𝑓(2+2ℎ)+ …+𝑓(2+(𝑛−1)ℎ)) =1 (𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 ((2)^2+(2+ℎ)^2+(2+2ℎ)^2+ …+(2+(𝑛−1)ℎ)^2 ) =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (█(2^2+(2^2+ℎ^2+4ℎ)+〖(2〗^2+(2ℎ)^2+8ℎ)+ …… @ …+(2^2+((𝑛−1)ℎ)^2+4(𝑛−1) ℎ) )) =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 [2^2+2^2+ … +2^2 ] + ℎ^2+(2ℎ)^2+ … +(𝑛−1)ℎ^2 + [4ℎ+8ℎ+ … +4(𝑛−1)ℎ] =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (〖𝑛(2)〗^2+[ℎ^2+(2)^2 . ℎ^2+ … +(𝑛−1)^2 ℎ^2 ] +[4ℎ+2×4ℎ+ … +(𝑛−1)×4ℎ] ) =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛(𝑛(2)^2+𝒉^2 [(1)^2+(2)^2+ …+(𝑛−1)^2 ] + 𝟒𝒉 [1+2+ …+(𝑛−1)]) =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (𝑛(2)^2+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)/6]+4ℎ[𝑛(𝑛 − 1)/2] ) 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 − 2 + 1) )/6 = (𝑛 (𝑛 − 1) (2𝑛 − 1) )/6 We know that 1+2+3+ ……+𝑛= (𝑛 (𝑛 + 1))/2 1+2+3+ ……+(𝑛−1) = ((𝑛 − 1) (𝑛 − 1 + 1))/2 = (𝑛 (𝑛 − 1) )/2 =(𝑙𝑖𝑚)┬(𝑛→∞) 1/𝑛 (4𝑛+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)]/6+2ℎ[𝑛(𝑛 − 1)] ) =(𝑙𝑖𝑚)┬(𝑛→∞) (4𝑛/𝑛+ℎ^2 [𝑛(𝑛 − 1)(2𝑛 − 1)/6𝑛]+2ℎ[𝑛(𝑛 − 1)/𝑛]) =(𝑙𝑖𝑚)┬(𝑛→∞) (4+ℎ^2 [(𝑛 − 1)(2𝑛 − 1)/6]+2ℎ[(𝑛 − 1)]) =(𝑙𝑖𝑚)┬(𝑛→∞) (4+(1/𝑛)^2 (𝑛 − 1)(2𝑛 − 1)/6+2(1/𝑛)(𝑛 − 1)) =(𝑙𝑖𝑚)┬(𝑛→∞) (4+1/𝑛^2 . (𝑛 − 1)(2𝑛 − 1)/6 +2(1 − 1/𝑛)) =(𝑙𝑖𝑚)┬(𝑛→∞) (4+ (1 − 1/𝑛)(2 − 1/𝑛)/6 +2(1 − 1/𝑛)) =4+ (1 − 1/∞)(2 − 1/∞)/6 +2(1 − 1/∞) =4+ (1 − 0)(2 − 0)/6 +2(1 −0) =4+ (1 . 2)/6 +2 .1 =4+ 1/3 +2 =6+ 1/3 = 𝟏𝟗/𝟑

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.