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0-1-t-2-1-2-t-6dx-




Question Number 146752 by Bens last updated on 15/Jul/21
  ∫_( 0) ^( 1)  t^2 + (1/2)t −6dx
$$ \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:{t}^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:−\mathrm{6}{dx}\:\: \\ $$
Answered by puissant last updated on 15/Jul/21
=t^2 ∫_0 ^1 1dx+(1/2)t∫_0 ^1 1dx−6∫_0 ^1 1dx  =t^2 +(1/2)t−6..
$$=\mathrm{t}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1dx}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1dx}−\mathrm{6}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1dx} \\ $$$$=\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}−\mathrm{6}.. \\ $$

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