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int.(x^2+[x])d(|3-x|) limit 0-6

ankit raj , 14 Years ago

Grade 12

1 Answers

Sher Mohammad

$\int_{0}^{6}x^2+[x])d(|3-x|) =\int_{0}^{3}x^2+[x])d(3-x)+\int_{3}^{6}x^2+[x])d(x-3)$
$=-\int_{0}^{3}x^2+[x] d(x)+\int_{3}^{6}x^2+[x] dx$
$=-\int_{0}^{3}x^2+-\int_{0}^{3}[x] d(x)+\int_{3}^{6}x^2+-\int_{3}^{6}[x] dx$
$\int_{0}^{3}[x] d(x)=\int_{0}^{1}1 d(x)+\int_{1}^{2}2 d(x)+\int_{2}^{3}3 d(x)$
similarly solve for other parts,

Thanks and Regards
Sher Mohammad
Askiitians Faculty,
B.tech, IIT Delhi

Last Activity: 11 Years ago

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