Oppgave 4.7 skrev:Regn ut integralene. Kontroller svarene ved å derivere.
a)
[tex]\int \left( x\cdot lnx \right)\rm{d}x[/tex]
[tex]u\prime = x \,\,\, u = \frac 12x^2 \\ \, \\ v = lnx \,\,\, v\prime = \frac 1x[/tex]
[tex]\int \left( x\cdot lnx \right)\rm{d}x = \frac 12x^2 \cdot lnx - \int \left(\frac 12x^{\cancel 2} \cdot \frac{1}{\cancel x}\right)\rm{d}x = \\ \, \\ \int \left( x\cdot lnx \right)\rm{d}x = \frac 12x^2 \cdot lnx - \int \left(\frac 12x\right)\rm{d}x = \\ \, \\ \int \left( x\cdot lnx \right)\rm{d}x = \frac 12x^2 \cdot lnx - \frac 14 x^2 + C = \underline{\underline{\frac 12x^2\left(lnx - \frac 12\right) + C}} [/tex]
Deriverer:
[tex]\left(\frac 12x^2\left(lnx - \frac 12\right) + C\right)\prime = \\ \, \\ \left(\frac12 x^2\cdot lnx - \frac 14 x^2 + C\right)\prime = \\ \, \\ \left( \left(\frac 12x^2\right)\prime \cdot lnx + \frac 12x^2 \cdot (lnx)\prime\right) - (\frac 14x^2)\prime + (C)\prime = \\ \, \\ x \cdot lnx + \frac 12x - \frac 12x + 0 = \\ \, \\ \underline{\underline{x\cdot lnx}}[/tex]
b)
[tex]u\prime = x^2 \,\,\, u = \frac 13 x^3 \\ \, \\ v\prime = \frac 1x \,\,\, v = lnx[/tex]
[tex]\int \left( x^2 \cdot lnx \right)\rm{d}x = \frac 13 x^3 \cdot ln x - \int \frac 13 x^{\cancel 3} \cdot \frac{1}{\cancel x}\rm{dx} = \\ \, \\ \int \left( x^2 \cdot lnx \right)\rm{d}x = \frac 13 x^3 \cdot ln x - \int \frac 13 x^{2} \rm{dx} = \\ \, \\ \int \left( x^2 \cdot lnx \right)\rm{d}x = \frac 13 x^3 \cdot ln x - \frac 19 x^{3}= \underline{\underline{\frac 13x^3\left(ln x - \frac 13\right) + C }}[/tex]
Jeg deriverer ikke dette uttrykket, for det er veldig likt som a). Jeg ser derfor at det stemmer.
c)
[tex]u\prime = \frac{1}{x^2} \,\,\, u = -\frac{1}{x} \\ \, \\ v\prime = \frac 1x \,\,\, v = lnx[/tex]
[tex]\int \left(\frac{1}{x^2} \cdot ln x\right)\rm{d}x = -\frac 1x \cdot lnx - \int \left( -\frac 1x \cdot \frac 1x\right)\rm{dx} \\ \, \\ \int \left(\frac{1}{x^2} \cdot ln x\right)\rm{d}x = -\frac 1x \cdot lnx + \int \left( x^{-2} \right)\rm{dx} \\ \, \\ \int \left(\frac{1}{x^2} \cdot ln x\right)\rm{d}x = -\frac 1x \cdot lnx - \frac 1x = \underline{\underline{ - \frac{lnx + 1}{x} +C}} [/tex]
d)
[tex]u\prime = e^x \,\,\, u = e^x \\ \, \\ v\prime = 1 \,\,\, v = x[/tex]
[tex]\int \left(x\cdot e^x \right)\rm{d}x = x\cdot e^x - \int \left(e^x\right) \rm{dx} \\ \, \\ \int \left(x\cdot e^x \right)\rm{d}x = x\cdot e^x - e^x = \underline{\underline{(x-1)e^x + C }}[/tex]
Deriverer:
[tex]\left( (x-1)\cdot e^x\right)\prime = \\ \, \\ (x-1)\prime \cdot e^x + (x-1) \cdot (e^x)\prime = \\ \, \\ e^x + xe^x - e^x = \underline{\underline{xe^x}}[/tex]
e)
[tex]u\prime = e^x \,\,\, u = e^x \\ \, \\ v\prime = 2 \,\,\, v = (2x-1)[/tex]
[tex]\int \left((2x+1)e^x \right)\rm{d}x = (2x+1)e^x - \int (2e^x)\rm{dx} \\ \, \\ \int \left((2x+1)e^x \right)\rm{d}x = (2x+1)e^x -2e^x = 2xe^x + e^x - 2e^x = 2xe^x - e^x = \underline{\underline{(2x-1)e^x + C}} [/tex]
Deriverer:
[tex]\left((2x-1)e^x\right)\prime = \\ \, \\ \left((2x-1)\prime \cdot e^x\right) + \left((2x-1)\cdot (e^x)\prime\right) = \\ \, \\ 2e^x + 2xe^x - e^x = \\ \, \\ 2xe^x + e^x = \underline{\underline{(2x+1)e^x}} [/tex]
f)
Skriver om integralet:
[tex]\int \left(\frac{ln x}{\sqrt x} \right)\rm{d}x =\int \left(x^{-\frac 12} \cdot lnx \right)\rm{d}x [/tex]
[tex]u\prime = x^{-\frac 12} \,\,\, u = 2x^{\frac 12} \\ \, \\ v\prime = \frac 1x \,\,\, v = lnx[/tex]
[tex]\int \left(x^{-\frac 12} \cdot lnx \right)\rm{d}x = 2x^{\frac 12} \cdot lnx - \int \left(2x^{\frac 12} \cdot \frac{1}{x}\right)\rm{dx}[/tex]
Vi kan endre det "andre" integralet slik:
[tex]\frac{2x^{\frac 12}}{x} = \frac{2}{x^{\frac 12}} = \underline{2x^{-\frac 12}}[/tex]
Dermed får vi:
[tex]\int \left(x^{-\frac 12} \cdot lnx \right)\rm{d}x = 2x^{\frac 12} \cdot lnx - \int \left(2x^{-\frac 12}\right)\rm{dx} \\ \, \\ \int \left(x^{-\frac 12} \cdot lnx \right)\rm{d}x = 2x^{\frac 12} \cdot lnx - \left(2 \cdot \frac{x^{-\frac 12 + 1}}{-\frac 12 + 1} \right) = 2x^{\frac 12}\cdot lnx - 4x^{\frac 12} = 2\sqrt x \cdot lnx - 4\sqrt x +C = \\ \, \\ \underline{\underline{2\sqrt x\left(lnx - 2\right) + C }}[/tex]
Vi deriverer for å kontrollere:
[tex]\left( 2\sqrt x \left(ln x-2\right) + C\right)\prime = \\ \, \\ \left(2x^{\frac 12} \cdot lnx - 4x^{\frac 12} + C\right)\prime = \\ \, \\ \left( (2x^{\frac 12})\prime \cdot lnx + 2x^{\frac 12} \cdot (lnx)\prime \right) - (4x^{\frac 12})\prime + (C)\prime = \\ \, \\ \left( 2\cdot \frac 12 x^{-\frac 12} \cdot lnx + 2x^{\frac 12} \cdot \frac 1x\right) - 4 \cdot \frac 12 x^{-\frac 12} + 0 = \\ \, \\ \left(x^{-\frac 12} \cdot lnx + \frac{2x^{\frac 12}}{x} \right) - 2x^{-\frac 12} = \\ \, \\ \left(\frac {lnx}{x^{\frac 12}} + \frac{2}{x^{1-\frac 12}\right) - \frac{2}{x^{\frac 12}} = \\ \, \\ \frac{lnx + 2 - 2}{x^{\frac 12}} = \underline{\underline{\frac{lnx}{\sqrt x}}} [/tex]