Integrasjonsregler

Kalkulus  >>  Integrasjon

 

f ,  g  og  u  er funkjsoner. a, b, r og  C  er konstanter.

Generelle integrasjonsregler

TYPE REGLER     
 EKSEMPEL  (Klikk for å se løsninger)
Linearitet \int (af + bg) dx =a \int f \hspace{0.5mm}dx + b \int g \hspace{0.5mm} dx \int (3 x^5 + 2 \sin x)dx = 3 \int x^5 dx + 2 \int \sin x \hspace{0.5mm} dx
Delvis integrasjon \int f\cdot g' \hspace{0.5mm} dx = fg- \int f'g \hspace{0.5mm} dx \int x \cdot (\sin x)' \hspace{0.5mm} dx = x \sin x - \int (x)' \sin x \hspace{0.5mm} dx
Substitusjon \int f(u) \cdot u' \hspace{0.5mm} dx = \int f(u) \hspace{0.5mm} du \int (\sin x) ^5 \cdot (\sin x)' \hspace{0.5mm} dx = \int u^5 \hspace{0.5mm} du,    der u = \sin x

Integralet av spesielle funksjoner

REGLER EKSEMPEL
\int a \hspace{0.5mm} dx = ax + C \int 5 \hspace{0.5mm} dx = 5x + C
\int x^r \hspace{0.5mm} dx = \frac{1}{r+1}x^{r+1} + C \int x^5 \hspace{0.5mm} dx = \frac{1}{5+1}x^{5+1} + C = \frac{1}{6}x^6 + C
\int \cos x \hspace{0.5mm} dx = \sin x + C
\int \sin x \hspace{0.5mm} dx = - \cos x + C
\int e^x \hspace{0.5mm} dx = e^x +C
\int x^{-1} \hspace{0.5mm} dx = \ln |x| +C
\int \frac{1}{\sqrt{1-x^2}} \hspace{0.5mm} dx = \arcsin x + C
\int \frac{1}{1+x^2} \hspace{0.5mm} dx = \arctan x + C

Eksempel i tabellen:

    \[ \int (3 x^5 + 2 \sin x)dx = 3 \int x^5 dx + 2 \int \sin x \hspace{0.5mm} dx = 3 \cdot \frac{1}{5+1}x^{5+1} + 2 (-\cos x) + C \]

    \[= 2x^6-2 \cos x + C\]

 

Flere eksempler

Mer om delvis integrasjon
Eksempel i tabellen:

    \[\int x \cdot \cos x \hspace{0.5mm} dx =\int x \cdot (\sin x)' \hspace{0.5mm} dx \hspace{2mm}\]

Vi lar f(x)=x  og  g(x)=\sin x

    \[\hspace{1mm}\int f\cdot g' \hspace{0.5mm} dx = fg- \int f'g \hspace{0.5mm} dx\]

    \[= x \sin x - \int (x)' \sin x \hspace{0.5mm} dx \hspace{1mm} = \hspace{1mm} x \sin x - \int \sin x \hspace{0.5mm} dx \hspace{1mm} = \hspace{1mm} x \sin x - (-\cos x) + C \]

    \[= x \sin x + \cos x + C \]

 
♦ Flere eksempler

    \[\int x \cdot e^{2x} \hspace{0.5mm} dx ,\]

Setter vi   u=x  og  v'=e^{2x}.    Da er u'=1   og   v=\frac{1}{2}e^{2x}

    \[\hspace{1mm}\int u\cdot v' \hspace{0.5mm} dx = uv- \int u'v \hspace{0.5mm} dx\]

    \[=x \cdot \frac{1}{2}e^{2x} - \int 1 \cdot \frac{1}{2}e^{2x}\hspace{0.5mm} dx \hspace{1mm} = \hspace{1mm} \frac{1}{2}x \cdot e^{2x}-\frac{1}{2} \int e^{2x}\hspace{0.5mm} dx \hspace{1mm} = \hspace{1mm} \frac{1}{2}x \cdot e^{2x}-\frac{1}{2} \cdot \frac{1}{2} e^{2x} + C \hspace{1mm} = \hspace{1mm} \frac{1}{2}x \cdot e^{2x}-\frac{1}{4} e^{2x} + C \]

Se video

    \[\int x^2 \cdot e^{2x} \hspace{0.5mm} dx ,\]

Setter vi   u=x^2  og  v'=e^{2x}.   Da er u'=2x   og   v=\frac{1}{2}e^{2x}

    \[\hspace{1mm}\int u\cdot v' \hspace{0.5mm} dx = uv- \int u'v \hspace{0.5mm} dx\]

    \[=x^2 \cdot \frac{1}{2}e^{2x} - \int 2x \cdot \frac{1}{2} \hspace{0.5mm} e^{2x} dx \hspace{1mm} = \hspace{1mm} \frac{1}{2}x^2 \cdot e^{2x} - \int x \cdot e^{2x}\hspace{0.5mm} dx \hspace{1mm} = \hspace{1mm} \frac{1}{2}x^2 e^{2x} - (\frac{1}{2} x \cdot e^{2x} - \frac{1}{4} e^{2x}) + C \]

    \[ = \hspace{1mm} \frac{1}{2}x^2 e^{2x} - \frac{1}{2} x \cdot e^{2x} + \frac{1}{4} e^{2x} + C \hspace{1mm} = \hspace{1mm} \frac{1}{4}e^{2x} (2x^2 - 2x+1) + C \]

Se video

Flere eksempler

Mer om substitusjon
Eksempel i tabellen:

    \[\int (\sin x) ^5 \cdot \cos x \hspace{0.5mm} dx =\int (\sin x) ^5 \cdot (\sin x)' \hspace{0.5mm} dx,\]

u = \sin x \hspace{2mm} \Leftrightarrow \hspace{2mm} u'=\frac{du}{dx}=\cos x \hspace{2mm} \Leftrightarrow \hspace{2mm} du= \cos x \hspace{0.5mm} dx

    \[\hspace{1mm}= \int u^5 \hspace{0.5mm} du \hspace{1mm} = \hspace{1mm} \frac{1}{5+1}u^{5+1}+C \hspace{1mm} = \frac{1}{6} u^6 + C \hspace{1mm} =\hspace{1mm} \frac{1}{6} (\sin x)^6 + C\]

♦ Flere eksempler

    \[\int x \cdot e^{-x^2} \hspace{0.5mm} dx = \int e^{-x^2} \cdot x \hspace{0.5mm} dx ,\]

u =-x^2 \hspace{2mm} \Leftrightarrow \hspace{2mm} u'=\frac{du}{dx}=-2x \hspace{2mm} \Leftrightarrow \hspace{2mm} du= -2x \hspace{0.5mm} dx \hspace{2mm} \Leftrightarrow \hspace{2mm} -\frac{1}{2} du = x dx

    \[\hspace{1mm}= \int e^u \cdot \hspace{0.5mm} (-\frac{1}{2}) du \hspace{1mm} = \hspace{1mm}- \frac{1}{2} \int e^u \hspace{0.5mm} du = -\frac{1}{2} e^u + C \hspace{1mm} = -\frac{1}{2} e^{-x^2} + C \]

Se video

    \[\int \dfrac{dx}{\sqrt{7x+5}} =\int \dfrac{1}{\sqrt{7x+5}} \hspace{0.5mm} dx,\]

u = 7x+5 \hspace{2mm} \Leftrightarrow \hspace{2mm} u'=\frac{du}{dx}=7 \hspace{2mm} \Leftrightarrow \hspace{2mm} du= 7 \hspace{0.5mm} dx \hspace{2mm} \Leftrightarrow \hspace{2mm} \dfrac{1}{7} du = dx

    \[\hspace{1mm}= \int \frac{1}{\sqrt{u}} \frac{1}{7} du \hspace{1mm} = \hspace{1mm} \frac{1}{7}\int \frac{1}{\sqrt{u}} \hspace{0.5mm} du \hspace{1mm} = \frac{1}{7}\int u^{-1/2} \hspace{0.5mm} du \hspace{1mm} =\hspace{1mm} \frac{1}{7} \cdot 2 u^{1/2} + C \hspace{1mm} =\hspace{1mm} \frac{2}{7} (7x+5)^{1/2} + C \]

    \[=\hspace{1mm} \frac{2}{7} \sqrt{7x+5} + C \]

Se video