MATHHX A
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1.2 Ubestemte integraler
I sidste afsnit lærte vi, at når vi har en funktion \(f\) og en stamfunktion \(F\), så kan samtlige stamfunktioner skrives på formen \(F(x)+c\). Det motiverer følgende definition:
-
Definition 1.2.1
Lad \(f\) være en funktion og \(F(x)\) en stamfunktion til \(f\). Det ubestemte integral af f er givet ved:
\[\int f(x)\,dx=F(x)+c,\]
hvor \(c\) er en arbitrær konstant.
Ordet arbitrær betyder i denne sammenhæng ”ikke videre bestemt”. På den måde kan man opfatte \(\int f(x)\,dx\) som at være en repræsentant for samtlige stamfunktioner til \(f\).
Integralet er altså ubestemt, fordi det ikke giver en konkret stamfunktion. Det er først når \(c\) får en værdi, at vi får en konkret stamfunktion. Notationen \(dx\) betyder at det er \(x\) som er variablen. Derfor læses
\(\int f(x)\,dx\) også nogle gange som det ubestemte integral af f af x med hensyn til x. Symbolet \(\int \) kaldes et integraltegn, \(f(x)\) kaldes integranden og \(c\)
kaldes integrationskonstanten . At regne det ubestemte integral kaldes også at integrere
-
Eksempel 1.2.1
Vi vil bestemme \(\int 5^x \,dx\) . Dvs. at vi skal finde en stamfunktion til \(5^x\) og så lægge \(c\) til. Vi slår stamfunktionen op i formelsamlingen, og får i alt:
\[\int 5^x \,dx=\frac {1}{\ln (5)}5^x+c\]
Øvelse 1.2.1
Med udgangspunkt i eksemplet oven over, skal du svare på følgende spørgsmål:
Løsning 1.2.1
-
a) \(5^x\)
-
b) \(x\)
-
c) \(c\)
Øvelse 1.2.2
Regn følgende ubestemte integraler
-
a) \(\int 2x \,dx\)
-
b) \(\int \ln (x) \,dx\)
Løsning 1.2.2
-
a) \(\int 2x \,dx = x^2+c\)
-
b) \(\int \ln (x) \,dx = x\ln (x)-x+c\)
Det er ikke altid variablen hedder \(x\). Det skal man dog ikke lade sig forvirre af. Det ændrer ikke noget i metoden.
Øvelse 1.2.3
Regn følgende ubestemte integraler
Løsning 1.2.3
-
a) \(\int 2t^4 \,dx= \frac {2}{5}t^5+c\)
-
b) \(\int \sqrt {u}+10 \,du=\frac {2}{3}u^{\frac {3}{2}}+10u+c\)
Der gælder følgende regler for ubestemte integraler:
Vi har allerede fundet stamfunktioner som har form som dem i sætningen (f.eks. i eksempel 1.1.2). Men indtil nu har vi været nødt til at gætte og så tjekke ved at
differentiere. Nu hvor vi har sætningen behøver vi ikke at tjekke at den fundne funktion rent faktisk er en stamfunktion.
-
Eksempel 1.2.2
Vi vil regne \(\int 5x^2\,dx\). Vi bruger reglen \(\int kf(x)\,dx=k\int f(x)\,dx\):
\[\int 5x^2\,dx=5\int x^2\,dx=5\frac {1}{3}x^3+c=\frac {5}{3}x^3+c\]
-
Eksempel 1.2.3
Vi vil regne \(\int x^2+e^x\,dx\). Vi bruger reglen \(\int (f(x)+g(x))\,dx=\int f(x)\,dx+\int g(x)\,dx\):
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int (x^2+e^x)\,dx &= \int x^2\,dx+\int e^x\,dx \\ &=\frac {1}{3}x^3+c_1+e^x+c_2 \\ &=\frac {1}{3}x^3+e^x+c
\end{align*}
Til sidst har vi samlet de to integrationskonstanter i en konstant \(c=c_1+c_2\). Normalt vil man dog springe mellemregningerne over og bare skrive:
\[\int (x^2+e^x)\,dx=\frac {1}{3}x^3+e^x+c.\]
Det betyder ikke at reglen er unødvendig. Vi har stadig brugt den – det fremgår bare ikke at den måde, vi har skrevet det op.
Øvelse 1.2.4
Bestem vha. sætning 1.2.1 følgende stamfunktioner. Skriv hvilken regel du bruger.
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a) \(\int x^2+2x\,dx\)
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b) \(\int 5e^x \,dx\)
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c) \(\int 4x^3+5^x-1 \,dx\)
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d) \(\int \frac {5}{x} \,dx\)
Løsning 1.2.4
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a) Vi bruger regel 2 og får \(\int x^2+2x\,dx=\frac {1}{3}x^3+x^2+c\)
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b) Vi bruger regel 1 og får \(\int 5e^x \,dx=5e^x+c\)
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c) Vi bruger først regel 3, så regel 2 og til sidste regel 1 og får \(\int 4x^3+5^x-1 \,dx=x^4+\frac {5^x}{\ln (5)}-x+c\)
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d) Vi skriver først integralet som \(\int 5\frac {1}{x} \,dx\). Så bruger vi regel 1 og får \(\int 5\frac {1}{x} \,dx=5\ln (|x|)+c\)