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Ett lite søtt dobbeltintegral krever eier;


[tex]I=\int_0^1 \, \int_{\sqrt[3]x}^{1}\frac{dydx}{1+y^4}[/tex]

Janhaa wrote:Ett lite søtt dobbeltintegral krever eier;


[tex]I=\int_0^1 \, \int_{\sqrt[3]x}^{1}\frac{dydx}{1+y^4}[/tex]

skifter grenser og får:

[tex]I=\int_0^1 \, \int_0^{y^3} \frac{1}{1+y^4}\, dx\,dy=\int_0^1 \frac{y^3}{1+y^4} \,dy[/tex]

setter [tex]u=1+y^4[/tex] som gir

[tex]I=\frac{1}{4}\int_1^2 \frac{1}{u}=\frac{1}{4}[\ln{u}]_1^2=\frac{1}{4}(\ln{2}-\ln{1})=\underline {\underline{\frac{1}{4}\ln{2}}}[/tex]

orjan_s wrote: skifter grenser og får:
[tex]I=\int_0^1 \, \int_0^{y^3} \frac{1}{1+y^4}\, dx\,dy=\int_0^1 \frac{y^3}{1+y^4} \,dy[/tex]
setter [tex]u=1+y^4[/tex] som gir
[tex]I=\int_1^2 \frac{1}{u}=[\ln{u}]_1^2=\ln{2}-\ln{1}=\underline {\underline{\ln{2}}}[/tex]

Blir vel;

[tex]I={1\over 4}\ln(2)[/tex]

ellers bra.

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