Integration Questions

Question Number 10876 by Saham last updated on 28/Feb/17

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \int_{\:\mathrm{x}} ^{\:\sqrt{\mathrm{x}}} \:\left(\mathrm{x}\:+\:\mathrm{y}^{\mathrm{5}} \right)\:\mathrm{dy}\:\mathrm{dx} \\$$

Answered by fariraihmudzengerere75@gmail.c last updated on 28/Feb/17

$${Answer}\:.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\int_{{x}} ^{\sqrt{{x}}} \:\:\left({x}+{y}^{\mathrm{5}} \right){dydx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left[\:\:{xy}\:+\frac{{y}^{\mathrm{6}} }{\mathrm{6}}+{c}\:\right]_{{x}} ^{\sqrt{{x}}} {dx}\: \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left[{x}×\sqrt{{x}\:}\:\:+\frac{\left(\sqrt{\left.{x}\right)^{\mathrm{6}} \:\:\:\:}\right.}{\mathrm{6}}−{x}^{\mathrm{2}\:\:} −\frac{{x}^{\mathrm{6}} }{\mathrm{6}}\:\:\:+{c}−{c}\right]{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[{x}^{\mathrm{3}/\mathrm{2}} +\:\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:−{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{6}} }{\mathrm{6}}\right]{dx} \\$$ $$=\left[\frac{{x}^{\mathrm{5}/\mathrm{2}} }{\mathrm{5}/\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}\:−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{x}^{\mathrm{7}} }{\mathrm{42}}\:\:+{c}−{c}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$ $$=\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{42}} \\$$ $$=\mathrm{0}\:.\mathrm{084}\:\:\mathrm{523}\:\:\mathrm{809} \\$$

Commented bySaham last updated on 28/Feb/17

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