MATHHX A
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1.3 Bestemte integraler
Jeg fortalte i indledning, at man kunne bruge integralregning til at bestemme arealer. Arealbestemmelse foregår vha. det som kaldes bestemte integraler, og det er emnet for dette afsnit. Først i et senere afsnit, vil vi se,
hvordan de bestemte integraler kan bruges til at finde arealer.
-
Definition 1.3.1
Lad \(f\) være en funktion og \(F\) en stamfunktion til \(f\). For to reelle tal \(a\) og \(b\) er det bestemte integral af \(f\) fra \(a\) til \(b\) givet ved:
\[\int _a^b f(x)\,dx=F(b)-F(a).\]
Tallet \(a\) kaldes den nedre grænse og tallet \(b\) kaldes den øvre grænse.
At regne et bestemt integral, med grænserne \(a\) og \(b\), kaldes også at integrere fra \(a\) til \(b\).
Når vi regner bestemte integraler, vil vi altid benytte skrivemåden:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _a^b f(x)\,dx & = \left [F(x)\right ]_a^b\\ & =F(b)-F(a)
\end{align*}
Skrivemåden gør det nemmere at overskue processen.
-
Eksempel 1.3.1
Vi vil bestemme det bestemte integral: \(\int _3^5 2x\,dx.\)
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _3^5 2x\,dx & = \left [x^2\right ]_3^5\\ & =5^2-3^2\\ & =16.
\end{align*}
Altså er
\[\int _3^5 2x\,dx=16.\]
Læg mærke til at det bestemte integral er et tal og ikke en funktion. Læg også mærke til at det er et bestemt tal. Det afhænger ikke at hvilken en stamfunktion man vælger (se næste øvelse).
Øvelse 1.3.1
Betragt det bestemte integral \(\int _3^5 2x\,dx\) fra eksempel 1.3.1.
-
a) Regn integralet, men brug nu stamfunktionen \(F(x)=x^2+7\). Får du samme tal som i eksemplet?
-
b) Forklar hvorfor det ikke spiller nogen rolle, hvilken stamfunktion man vælger når man regner bestemte integraler.
Øvelse 1.3.2
Regn følgende bestemte integraler
-
a) \(\int _2^3 3x^2\,dx\)
-
b) \(\int _{-2}^5 (6t-2)\,dt\)
-
c) \(\int _0^1 e^x\,dx\)
-
d) \(\int _{-e}^{-1} \frac {1}{u}\,du\)
-
e) \(\int _{2}^{-1} 5 \,dx\)
Løsning 1.3.2
-
a) \(\int _2^3 3x^2\,dx=19\)
-
b) \(\int _{-2}^5 (6t-2)\,dt=49\)
-
c) \(\int _0^1 e^x\,dx=e-1\approx 1{,}72\)
-
d) \(\int _{-e}^{-1} \frac {1}{u}\,du=-1\)
-
e) \(\int _{2}^{-1} 5 \,dx=-15\)
Ligesom ved ubestemte integraler gælder der nogle regneregler for bestemte integraler:
-
Sætning 1.3.1
Lad \(f\) og \(g\) være kontinuerte funktioner, \(a\), \(b\) og \(c\) reelle tal, og \(k\) en
konstant. Så gælder følgende regneregler:
-
1. \(\int _a^b k\cdot f(x)\,dx=k\cdot \int _a^b f(x)\,dx\)
-
2. \(\int _a^b \left ( f(x)+g(x)\right ) \,dx=\int _a^b f(x)\,dx+\int _a^b g(x)\,dx\)
-
3. \(\int _a^b \left ( f(x)-g(x)\right ) \,dx=\int _a^b f(x)\,dx-\int _a^b g(x)\,dx\)
-
4. \(\int _a^b f(x)\,dx=\int _a^c f(x)\,dx + \int _c^b f(x)\,dx\) (Indskudsreglen)
De to første regler er fuldstændig tilsvarende til dem vi så for ubestemte integraler.
-
Eksempel 1.3.2
Regel 1:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _{-2}^3 10x\,dx = & 10 \int _{-2}^3 x\,dx\\ = & 10\left [\frac {1}{2}x^2\right ]_{-2}^3\\ = & 10(\frac {1}{2} 3^2-\frac {1}{2} (-2)^2)\\ = & 25
\end{align*}
Regel 2:
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _1^3 \left (6x+2\right ) \,dx & = \int _1^3 6x\,dx+\int _1^3 2\,dx\\ & = \left [3x^2\right ]_1^3 + \left [ 2x \right ]_1^3\\ & = 3\cdot 3^2-3\cdot 1^2 + 2\cdot 3 -2 \cdot 1\\
& = 27-3+6-2\\ & = 28
\end{align*}
Bare fordi man kan bruge en regel, betyder det ikke altid at det er smart at bruge den. Vi vil nu regne det sidste integral fra ovenstående eksempel uden at bruge regel 2.
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _1^3 \left (6x+2\right ) \,dx & = \left [3x^2 + 2x \right ]_1^3\\ & = 3\cdot 3^2+ 2\cdot 3 - (3\cdot 1^2+ 2\cdot 1)\\ & = 27+6-3-2\\ & = 28
\end{align*}
Var det ikke nemmere? Mens det næsten altid er smart at trække konstanter ud foran integraltegnet (regel 1), er det sjældent smart at dele summer/differenser op (regel 2). Ligesom man kan tage konstanter ud foran integraltegnet,
kan man også tage dem ud foran de firkantede parenteser:
\[\left [k\cdot F(x)\right ]_a^b=k\mleft [F(x)\mright ]_a^b\]
Igen en nyttig regel som kunne have gjort udregninger i eksempel 1.3.2 en anelse nemmere:
-
Eksempel 1.3.3
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _{-2}^3 10x\,dx = & 10 \int _{-2}^3 x\,dx \\ = & 10\left [\frac {1}{2}x^2\right ]_{-2}^3\\ = & 10\cdot \textcolor {red}{\frac {1}{2} }\left [x^2\right ]_{-2}^3 &&
\text {(Rykker konstant ud)}\\ = & 10\cdot \frac {1}{2}(3^2 - (-2)^2)\\ = & 25
\end{align*}
Indskudsreglen (sætning 1.3.1 punkt 4) er nyttig til integraler, hvor integranden er stykkevist defineret:
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Eksempel 1.3.4
Lad funktionen \(f\) være defineret ved:
\[ f(x) = \begin {cases} x^2 & x< 1 \\ \frac {1}{x} & x \geq 1 \end {cases} \]
Vi vil regne integralet
\[\int _0^2 f(x)\, dx.\]
Da \(f\) er stykkevist defineret kan vi ikke finde en samlet stamfunktion, så vi bruger indskudsreglen til at dele den op, der hvor forskriften skifter (i \(x=1\)):
\(\seteqnumber{0}{1.}{0}\)
\begin{align*}
\int _0^2 f(x)\, dx = & \int _0^1 f(x)\, dx + \int _1^2 f(x)\, dx && (\text {Indskudsregel med }c=1)\\ = & \int _0^1 x^2\, dx +\int _1^2 \frac {1}{x} \, dx\\ = & \left
[\frac {1}{3}x^3\right ]_0^1+\big [\ln (|x|)\big ]_1^2 \\ = & \frac {1}{3}(1^3-0^3) + \ln |2|-\ln |1| && \text {(Rykket konstant ud)}\\ = & \frac {1}{3} + \ln (2)\\ = &
1{,}03
\end{align*}
Altså er
\[\int _0^2 f(x)\, dx = 1{,}03\]
Øvelse 1.3.3
Betragt funktionen \(f\) med forskriften:
\[f(x) = \begin {cases} e^x & x< 0 \\ x +1 & x \geq 0 \end {cases}\]
Ekstra
Sætning 1.3.1 indeholder de vigtigste regler for bestemte integraler, men for at gøre listen mere komplet, udvider vi med følgende regler:
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1. \(\int _a^b f(x)\,dx=-\int _b^a f(x)\,dx\)
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2. \(\int _a^a f(x)\,dx=0\)
Øvelse 1.3.4
Lad \(f\) være en funktion med en hemmelig forskrift. Antag at:
\[\int _2^7 f(x) \, dx = 10 \quad \text {og} \quad \int _4^7 f(x) \, dx =3\]
Bestem
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a) \(\int _2^7 3\cdot f(x) \, dx\)
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b) \(\int _4^7 f(x)-2 \, dx\)
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c) \(\int _7^2 f(x)\, dx\)
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d) \(\int _5^5 f(x)\, dx\)
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e) \(\int _2^4 f(x)\, dx\)
Løsning 1.3.4
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a) \(\int _2^7 3\cdot f(x) \, dx = 30\)
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b) \(\int _4^7 f(x)-2 \, dx = -3\)
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c) \(\int _7^2 f(x)\, dx=-10\)
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d) \(\int _5^5 f(x)\, dx=0\)
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e) \(\int _2^4 f(x)\, dx= 7\)