Posted: 02/08-2011 18:03
Skuffende lite aktivitet i denne tråden (Selv om janhaa gjør en stor innsats), og egentlig ellers i forumet. Men det er vel ingen som kan klandres for =(
Hva vil dere si er artige problem ?
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Løsningen på integralet som stod uløst fra forrige side =)
[tex] I_4 = \int\limits_0^\infty {\frac{{{x^{29}}}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}}\:dx} [/tex]
[tex] I_4 = \int\limits_0^\infty {\frac{x\cdot (x^2)^{14}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}} \:dx} [/tex]
[tex] y=x^2 \; \; , \; \; dy = 2x \, dx [/tex]
[tex] I_4 = \frac{1}{2} \int\limits_0^\infty {\frac{ (y)^{14}}{{{{\left( {5{y} + 49} \right)}^{17}}}} \:dy} [/tex]
[tex] y=\frac{49}{5}z \; \; , \;\; dy = \frac{49}{5} dz [/tex]
[tex] I_4 = \frac{1}{2}\int\limits_0^\infty {\frac{ (\frac{49}{5}z)^{14}}{{{{\left( {49 z + 49} \right)}^{17}}}} \frac{49}{5}\:dz} [/tex]
[tex] I_4 = \frac{1}{2} \cdot \frac{49}{5} \cdot \frac{49^{14}}{5^{14}} \cdot \frac{1}{49^{17}}\int\limits_0^\infty {\frac{ z^{14}}{{{{\left( {z + 1} \right)}^{17}}}} \: dz} [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{\infty} \left( \frac{z}{z+1} \right)^{14}\cdot \frac{1}{(z+1)^3} \: dz [/tex]
[tex] y=\frac{z}{z+1} \; \; , \; \; dy = \frac{1}{(z+1)^2} dx \; \; , \; \; \frac{1}{z+1}=(1-y)[/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{1} y^{14}(1-y) \: dy [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{1} y^{14}-y^{15}\:dy [/tex]
[tex] \int_{0}^{1} x^n dx \, = \, \left[ \frac{1}{n+1}{x^{n+1}}\right]_{0}^{1} \, = \, \frac{1}{n+1} [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\left( \frac{1}{15}-\frac{1}{16}\right) [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}\cdot 15 \cdot 16}[/tex]
[tex] I_4 = \int\limits_0^\infty {\frac{{{x^{29}}}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}} \: dx} \, = \, \frac{14!}{2 \cdot 49^2 \cdot 5^{15}\cdot 16!} [/tex]
Q.E.D
Hva vil dere si er artige problem ?
---------------------------------------------
Løsningen på integralet som stod uløst fra forrige side =)
[tex] I_4 = \int\limits_0^\infty {\frac{{{x^{29}}}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}}\:dx} [/tex]
[tex] I_4 = \int\limits_0^\infty {\frac{x\cdot (x^2)^{14}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}} \:dx} [/tex]
[tex] y=x^2 \; \; , \; \; dy = 2x \, dx [/tex]
[tex] I_4 = \frac{1}{2} \int\limits_0^\infty {\frac{ (y)^{14}}{{{{\left( {5{y} + 49} \right)}^{17}}}} \:dy} [/tex]
[tex] y=\frac{49}{5}z \; \; , \;\; dy = \frac{49}{5} dz [/tex]
[tex] I_4 = \frac{1}{2}\int\limits_0^\infty {\frac{ (\frac{49}{5}z)^{14}}{{{{\left( {49 z + 49} \right)}^{17}}}} \frac{49}{5}\:dz} [/tex]
[tex] I_4 = \frac{1}{2} \cdot \frac{49}{5} \cdot \frac{49^{14}}{5^{14}} \cdot \frac{1}{49^{17}}\int\limits_0^\infty {\frac{ z^{14}}{{{{\left( {z + 1} \right)}^{17}}}} \: dz} [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{\infty} \left( \frac{z}{z+1} \right)^{14}\cdot \frac{1}{(z+1)^3} \: dz [/tex]
[tex] y=\frac{z}{z+1} \; \; , \; \; dy = \frac{1}{(z+1)^2} dx \; \; , \; \; \frac{1}{z+1}=(1-y)[/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{1} y^{14}(1-y) \: dy [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\int_{0}^{1} y^{14}-y^{15}\:dy [/tex]
[tex] \int_{0}^{1} x^n dx \, = \, \left[ \frac{1}{n+1}{x^{n+1}}\right]_{0}^{1} \, = \, \frac{1}{n+1} [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}}\left( \frac{1}{15}-\frac{1}{16}\right) [/tex]
[tex] I_4 = \frac{1}{2 \cdot 49^2 \cdot 5^{15}\cdot 15 \cdot 16}[/tex]
[tex] I_4 = \int\limits_0^\infty {\frac{{{x^{29}}}}{{{{\left( {5{x^2} + 49} \right)}^{17}}}} \: dx} \, = \, \frac{14!}{2 \cdot 49^2 \cdot 5^{15}\cdot 16!} [/tex]
Q.E.D