MATHHX A
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1.6 Integraler i GeoGebra
Man finder en stamfunktion med kommandoen :
\[\verb |Integral( <Funktion> )|\]
-
Eksempel 1.6.1
Vi vil bestemme en stamfunktion \(F\) til \(f(x)=2x\) i Geogebra. I et algebravindue skriver vi: \(\verb |Integral(2x)|\):
og vi kan se at \(F(x)=x^2\) er en stamfunktion til \(f\) (vi skriver \(F\) selvom Geogebra kalder stamfunktionen \(f\))
Øvelse 1.6.1
Lad \(f(x)=x\cdot e^x\).
Ubestemte integraler bestemmer man også med kommandoen \(\verb |Integral(funktion)|\), men man skal indtaste den i CAS:
-
Eksempel 1.6.2
Vi vil bestemme \(\int 2x \, dx\) i Geogebra. Vi åbner CAS og skriver: \(\verb |Integral(2x)|\):
Vi kan se at Geogebra kalder integrationskonstanten \(c_1\). Vi plejer at kalde den for \(c\) vores facit er \(\int 2x \, dx=x^2+c\)
Øvelse 1.6.2
Lad \(f(x)= \sqrt {x+1} \).
Vil man bestemme en bestemt stamfunktion (f.eks. gennem et punkt), så finder man først det ubestemte integral, og derefter bruger man kommandoen \(\verb |Beregn(ligning med x)|\) til at finde integrationskonstanten.
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Eksempel 1.6.3
Vi vil bestemme den stamfunktion \(F\) til \(f(x)=2x\), som går igennem punktet \((4,5)\).
Vi skal igen bruge CAS. Vi starter med regne det ubestemte integral, men denne gang vil vi gerne kalder resultatet \(F\), så vi skriver: \(\verb |F(x):=Integral(2x)|\).
Vi skal nu finde ud af hvad integrationskonstanten skal være så vi skriver: \(\verb |Beregn(F(4)=5)|\):
Vi ser at \(c_1=-11\), så vores facit er:
\(F(x)=x^2-11\)
Øvelse 1.6.3
Lad \(f(x)=x+\frac {1}{x}\).
Bestemte integraler
Bestemte integraler regner man med kommandoen:
\[\verb |Integral(funktion, Start x-Værdi, Slut x-Værdi)|\]
-
Eksempel 1.6.4
Vi vil regne det bestemte integral: \(\int _3^5 2x\,dx.\) I et algebra vindue skriver vi: \(\verb |Integral(2x,3,5)|\):
Altså er
\[\int _3^5 2x\,dx=16.\]
Øvelse 1.6.4
Regn i GeoGebra:
Areal under graf
Da arealer under grafer er givet ved bestemte integraler er der ikke rigtig noget ny her i forhold til Geogebra. Men okay her er et eksempel alligevel:
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Eksempel 1.6.5
Lad \(f(x)=0{,}1e^x\). Vi vil bestemme arealet af området M, afgrænset af førsteaksen, andenaksen, funktionen \(f\) og linjen med ligningen \(x=3\).
Vi skriver: \(\verb |Integral(0,1e^x,0,3)|\):
Vi kan se at \(A= 1{,}91\). Vi kan endda se selve arealet i tegneblokken til højre.
Øvelse 1.6.5
Lad \(f(x)=x^3-e^x\). Lad \(A\) betegne arealet af punktmængden afgrænset af førsteaksen, andenaksen, \(f\) og linjen \(x=1\).
Øvelse 1.6.6
Lad \(f(x)=x^3-7x^2+14x-8\). Grafen for \(f\) afgrænser sammen med førsteaksen en punktmængde som vist her:
Måske er det mere interessant at se et eksempel på hvordan man bestemmer en ukendt konstant i forbindelse med et integral.
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Eksempel 1.6.6
Betragt arealet \(A\) af området afgrænset førsteaksen, funktionen \(f(x))=\frac {1}{4}x^2\), linjen \(x=1\) og linjen \(x=b\). Vi vil bestemme \(b\) således at \(A=2\).
Vi åbner CAS og skriver: \(\verb |A:=Integral(1/4x^2,1,b)|\) og får:
Vi kan nu finde \(b\) ved at løse ligningen \(A=2\). Vi skriver: \(\verb |Beregn(A=2)|\)
Vi vil gerne have decimaltal, så vi trykker krøllet lighedstegn:
og vi får facit:
Altså \(b=2{,}92\)
Øvelse 1.6.7
Lad \(f(x)=\sqrt {x^2+k}\). Lad \(A\) betegne arealet afgrænset af førsteaksen, andenaksen, \(f\) og linjen \(x=1\)
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a) Hvis GeoGebra-kommandoen Beregn ikke virker, findes der en anden kommando man kan bruge. Hvad er det for en kommando? og hvad
er forskellen på de to kommandoer?
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b) Bestem \(k\) så \(A=1\).
Løsning 1.6.7
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a) Kommandoen hedder NBeregn. Forskellen er at Beregn forsøger at finde en eksakt løsning (f.eks. på form som en brøk eller
en kvadratrod), mens NBeregn finder en decimaltalsløsning (f.eks. ”\(2{,}45\)). Kommandoen Beregn er finere, hvis man er en rigtig matematiker, men det vi jo ikke, så bare brug NBeregn.
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b) \(k=0{,}69\)
Areal mellem to grafer
Arealet mellem to grafer findes med kommandoen:
\[\small \verb |IntegralMellem(funktion, Funktion, Start x-Værdi, Slut x-Værdi)|\]
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Eksempel 1.6.7
Vi vil bestemme arealet \(A\) af området \(M\) mellem graferne for funktionerne \(f(x)=-x^2+5x-3\) og \(g(x)=-x+2\).
Vi starter med at tegne funktionerne:
Vi finder skæringspunkterne med skæringsværktøjet eller skæringskommandoen:
Vi kan se at \(x\)-koordinaterne til skæringspunkterne er \(1\) og \(5\). Dvs. vi har de grænser vi skal bruge og vi kan regne arealet med et integral.
Vi skriver: \(\verb |IntegralMellem(f,g,1,5)|\)
Vi kan se på tegneblokken at vi har fat i det rigtige areal, og at det har værdien \(10{,}67\).
Øvelse 1.6.8
Lad \(f(x)=2\cdot 0{,}5^{x-2}\) og betragt punktmængden \(M\):