Hjelp til dobbeltintegral
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Dette integralet løses lett ved å bytte integrasjonsrekkefølgen:
[tex]\int_0^1 \: \int_{x^{1/3}}^1 \: \sqrt{1 \:-\: y^4} \: dy\,dx[/tex]
[tex]=\; \int_0^1 \: \int_0^{y^3} \: \sqrt{1 \:-\: y^4} \: dx\,dy[/tex]
[tex]=\; \int_0^1 \: \Big[\, x \, \sqrt{1 \:-\: y^4} \, \Big]_0^{y^3} \,dy [/tex]
[tex]=\; \int_0^1 \: y^3 \, \sqrt{1 \:-\: y^4} \,dy [/tex]
[tex]=\; \int_1^0 \: -\frac{u^2}{2} \,du\;\;[/tex] (Anvender substitusjonen [tex]u = \sqrt{1 \:-\: y^4}[/tex] )
[tex]= \; \Big[ \, -\frac{u^3}{6} \, \Big]_1^0\;\;[/tex]
[tex]=\; \underline{\underline{\frac{1}{6}}}\,. [/tex]
[tex]\int_0^1 \: \int_{x^{1/3}}^1 \: \sqrt{1 \:-\: y^4} \: dy\,dx[/tex]
[tex]=\; \int_0^1 \: \int_0^{y^3} \: \sqrt{1 \:-\: y^4} \: dx\,dy[/tex]
[tex]=\; \int_0^1 \: \Big[\, x \, \sqrt{1 \:-\: y^4} \, \Big]_0^{y^3} \,dy [/tex]
[tex]=\; \int_0^1 \: y^3 \, \sqrt{1 \:-\: y^4} \,dy [/tex]
[tex]=\; \int_1^0 \: -\frac{u^2}{2} \,du\;\;[/tex] (Anvender substitusjonen [tex]u = \sqrt{1 \:-\: y^4}[/tex] )
[tex]= \; \Big[ \, -\frac{u^3}{6} \, \Big]_1^0\;\;[/tex]
[tex]=\; \underline{\underline{\frac{1}{6}}}\,. [/tex]