espen180 wrote:I got carried away...
[tex]a) \\ f(e^{2x})=6x-3e^{4x}[/tex]
[tex]b) \\ \int_1^{f(e)} f(u) \rm{d}u[/tex]
f er lik i a og b.
a)
[tex]u = (e^x)^2 \\ \, \\ \sqrt u = e^x \\ \, \\ \frac 12 \ln u = x\ln e \\ \, \\ x = \frac 12\ln u[/tex]
Dermed;
[tex]6\frac 12 \ln u-3e^{\frac 12\ln u \cdot 4} \\ \, \\ \Rightarrow \; 3\ln u -3e^{2\ln u} \\ \, \\ \Rightarrow \; 3\ln u -3(e^{\ln u})^2 \\ \, \\ \Rightarrow \; 3\ln u - 3u^2[/tex]
[tex]f(u) = 3\ln u - 3u^2[/tex]
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Prøver:
[tex]f(e^{2x}) = 3\ln \left(e^{2x}\right) - 3(e^{2x})^2 = 3\cdot 2x \ln e - 3e^{4x} = \underline{6x - 3e^{4x}}[/tex]
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b)
Regner ut øvre grense:
[tex]f(e) = 3\ln e - 3e^2 = \underline{3-3e^2}[/tex]
[tex]1\;>\; 3-3e^2[/tex]
Dropper grensene i første omgang.
[tex]\int \left( 3\ln u -3u^2\right)\rm{d}u = \int 3\ln u \rm{d}u - 3\int u^2 \rm{d}u =[/tex]
[tex]\int \left( 3\ln u -3u^2\right)\rm{d}u = 3u \ln u - \int 3\cancel u \frac{1}{\cancel u}\rm{d}u - 3 \int u^2 \rm{d}u =[/tex]
[tex]\int \left( 3\ln u -3u^2\right)\rm{d}u = 3u\ln u - 3u - u^3 + C[/tex]
[tex]F(u) = u\left(3\ln |u| - 3 - u^2\right) + C[/tex]
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Siden [tex]a<b[/tex] i [tex]\int_b^a f(u)\rm{d}u \Rightarrow -\int_a^b f(u)\rm{d}u[/tex]
[tex]-\left[3u\ln|u|-3u - u^3\right]_{3-3e^2}^{1} = \left[u^3+3u-3u\ln|u|\right]_{3-3e^2}^{1} [/tex]
Jeg tror ikke jeg orker den ass.
