Integraler med substitusjon S2 V26
Bestem integralene
Oppgave
- $$\int_0^2 \left(e^{2x} + x\right) \, \mathrm{d}x$$
- $$\int \frac{(\ln x)^2}{x} \, \mathrm{d}x$$
Fasit
a) \(\frac{e^{4}+3}{2}\)
b) \(\frac{(\ln x)^{3}}{3}+C\)
Løsningsforslag
a
\[\begin{aligned} \int_{0}^{2} \left( e^{2x}+x \right) \, \mathrm{d}x &= \left[ \frac{1}{2}e^{2x}+ \frac{1}{2}x^{2} \right]_{0}^{2} \\ &= \frac{1}{2} \left[ e^{2x}+x^{2} \right]_{0}^{2} \\ &= \frac{1}{2}\left( \left( e^{2 \cdot 2}+2^{2} \right) - \left( e^{2 \cdot 0} + 0 ^{2} \right) \right) \\ &= \frac{1}{2} \left( \left( e^{4}+4 \right) - \left( 1 \right) \right) \\ &= \frac{1}{2}\left( e^{4} +3 \right) = \underline{\underline{ \frac{e^{4}+3}{2} }} \end{aligned} \]
b
Vi lar \(u= \ln x\) og gjør variabelskifte.
\[\textcolor{steelblue}{u} = \textcolor{steelblue}{\ln x} \implies \frac{du}{dx}=\frac{1}{x}\implies \textcolor{seagreen}{du \cdot x} = \textcolor{seagreen}{dx} \]
\[\begin{aligned} \int \frac{(\textcolor{steelblue}{\ln x)}^2}{x}\,\textcolor{seagreen}{\mathrm{d}x} &= \int \frac{\textcolor{steelblue}{u}^{2}}{x} \, \textcolor{seagreen}{du \cdot x} \\ &= \int u^{2} \, \mathrm{d}u \\ &= \frac{1}{3}u^{3}+C \\ &= \underline{\underline{ \frac{(\ln x)^{3}}{3}+C }} \end{aligned} \]