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Question Number 80951 by TawaTawa last updated on 08/Feb/20

Commented by TawaTawa last updated on 08/Feb/20

Please help.

$$\mathrm{Please}\:\mathrm{help}. \\ $$

Commented by john santu last updated on 08/Feb/20

(1) ∫_1 ^2  ∫_0 ^x^2   x^2  dydx =   ∫_1 ^2  x^2 y∣_0 ^x^2    dx = ∫_1 ^2  (x^4 )dx   = (1/5)x^5  ∣_1 ^2  = ((31)/5)

$$\left(\mathrm{1}\right)\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:{x}^{\mathrm{2}} \:{dydx}\:=\: \\ $$$$\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:{x}^{\mathrm{2}} {y}\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\mid}}\:\:{dx}\:=\:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\int}}\:\left({x}^{\mathrm{4}} \right){dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{5}}{x}^{\mathrm{5}} \:\underset{\mathrm{1}} {\overset{\mathrm{2}} {\mid}}\:=\:\frac{\mathrm{31}}{\mathrm{5}} \\ $$

Commented by TawaTawa last updated on 08/Feb/20

God bless you sir. I appreciate

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

Answered by mr W last updated on 08/Feb/20

(1)  ∫_1 ^2 ∫_0 ^x^2  x^2 dydx  =∫_1 ^2 x^2 ∫_0 ^x^2  dydx  =∫_1 ^2 x^2 [y]_0 ^x^2  dx  =∫_1 ^2 x^2 x^2 dx  =∫_1 ^2 x^4 dx  =(1/5)[x^5 ]_1 ^2   =(1/5)(32−1)  =((31)/5)    (2)  ∫_0 ^1 ∫_0 ^( y) (x^2 +y^2 )dxdy  =∫_0 ^1 [(x^3 /3)+y^2 x]_0 ^y dy  =∫_0 ^1 ((y^3 /3)+y^2 y)dy  =∫_0 ^1 (((4y^3 )/3))dy  =[(y^4 /3)]_0 ^1   =(1/3)

$$\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{{x}^{\mathrm{2}} } {x}^{\mathrm{2}} {dydx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} \int_{\mathrm{0}} ^{{x}^{\mathrm{2}} } {dydx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} \left[{y}\right]_{\mathrm{0}} ^{{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} {x}^{\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{4}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left[{x}^{\mathrm{5}} \right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{32}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{31}}{\mathrm{5}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\:{y}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{y}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{{y}} {dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{y}^{\mathrm{3}} }{\mathrm{3}}+{y}^{\mathrm{2}} {y}\right){dy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{4}{y}^{\mathrm{3}} }{\mathrm{3}}\right){dy} \\ $$$$=\left[\frac{{y}^{\mathrm{4}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by TawaTawa last updated on 08/Feb/20

God bless you sir. I appreciate.

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

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