Evaluate
∫ (x ^{ 3 } + 1) dx from -2 to 2
Last updated at Oct. 17, 2019 by Teachoo
Transcript
Question 17 โซ (x3 + 1) dx from -2 to 2 This is of form โซ a -a f (x) dx And we now that โซ_(โ๐)๐ ๐(๐ฅ)๐๐ฅ=0,ใ if f(โ๐ฅ)=โ๐(๐ฅ) And โซ_(โ๐)^๐๐(๐ฅ)๐๐ฅ=2โซ_0^๐๐(๐ฅ)๐๐ฅ , if f(โ๐ฅ)=๐(๐ฅ) Now, โซ_(โ2)^2(๐ฅ^3+1)๐๐ฅ = โซ_(โ2)^2 ๐ฅ^3 ๐๐ฅใ + โซ_(โ2)^21๐๐ฅ Since ใ(โ๐ฅ)ใ^3=โ๐ฅ^3 So, โซ_(โ๐)^๐โใ๐ฅ^3 ๐๐ฅ=0,ใ And 1 is constant, so f(โx) = f(x) โซ_(โ๐)^๐โ1๐๐ฅ=2โซ_0^๐โ๐๐ฅ = 0 + 2โซ_0^2โ1๐๐ฅ = 2โซ_0^2โ๐๐ฅ = 2 ใ[๐ฅ]ใ_0^2 = 2 (2 โ 0) = 4
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CBSE Class 12 Sample Paper for 2020 Boards
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