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

∫_( 0) ^( 1) ∫_( x) ^( (√x))  (x + y^5 ) dy dx

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

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

Answer . ∫_0 ^1  ∫_x ^(√x)   (x+y^5 )dydx  =∫_0 ^1  [  xy +(y^6 /6)+c ]_x ^(√x) dx   =∫_0 ^1  [x×(√(x ))  +((((√(x)^6     )))/6)−x^(2  ) −(x^6 /6)   +c−c]dx  =∫_0 ^1 [x^(3/2) + (x^3 /6) −x^2 −(x^6 /6)]dx  =[(x^(5/2) /(5/2))+(x^4 /(24)) −(x^3 /3) −(x^7 /(42))  +c−c]_0 ^1   =(2/5)+(1/(24))−(1/3)−(1/(42))  =0 .084  523  809

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

Commented by Saham last updated on 28/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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