Question Number 201702 by hardmath last updated on 10/Dec/23
$$\mathrm{Find}: \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{3}} \:\mathrm{dx}\:\int_{\boldsymbol{\mathrm{x}}} ^{\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \:\left(\mathrm{x}\:−\:\mathrm{y}\right)\:\mathrm{dy} \\ $$
Commented by mr W last updated on 11/Dec/23
$${take}\:{care}!\:{i}\:{think}\:{you}\:{want}\:{to}\:{mean} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \int_{{x}} ^{{x}^{\mathrm{3}} } \left({x}−{y}\right){dydx} \\ $$
Answered by mr W last updated on 11/Dec/23
![∫_1 ^3 ∫_x ^x^3 (x−y)dydx =∫_1 ^3 [∫_x^3 ^x (x−y)d(x−y)]dx =∫_1 ^3 [(((x−y)^2 )/2)]_x^3 ^x dx =−∫_1 ^3 (((x−x^3 )^2 )/2)dx =−(1/2)∫_1 ^3 (x^2 −2x^4 +x^6 )dx =−(1/2)[(x^3 /3)−((2x^5 )/5)+(x^7 /7)]_1 ^3 =−(1/2)(((3^3 −1^3 )/3)−((2(3^5 −1^5 ))/5)+((3^7 −1^7 )/7)) =−(1/2)(((26)/3)−((2×242)/5)+((2186)/7)) =−((11768)/(105))](https://www.tinkutara.com/question/Q201720.png)
$$\int_{\mathrm{1}} ^{\mathrm{3}} \int_{{x}} ^{{x}^{\mathrm{3}} } \left({x}−{y}\right){dydx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \left[\int_{{x}^{\mathrm{3}} } ^{{x}} \left({x}−{y}\right){d}\left({x}−{y}\right)\right]{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\left({x}−{y}\right)^{\mathrm{2}} }{\mathrm{2}}\right]_{{x}^{\mathrm{3}} } ^{{x}} {dx} \\ $$$$=−\int_{\mathrm{1}} ^{\mathrm{3}} \frac{\left({x}−{x}^{\mathrm{3}} \right)^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} \left({x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{4}} +{x}^{\mathrm{6}} \right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}}\right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}\left(\mathrm{3}^{\mathrm{5}} −\mathrm{1}^{\mathrm{5}} \right)}{\mathrm{5}}+\frac{\mathrm{3}^{\mathrm{7}} −\mathrm{1}^{\mathrm{7}} }{\mathrm{7}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{26}}{\mathrm{3}}−\frac{\mathrm{2}×\mathrm{242}}{\mathrm{5}}+\frac{\mathrm{2186}}{\mathrm{7}}\right) \\ $$$$=−\frac{\mathrm{11768}}{\mathrm{105}} \\ $$
Commented by hardmath last updated on 11/Dec/23
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor} \\ $$