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How to Mathe Abi

05) Integrale (Analysis)

RK
by Rick K.

Definiere den Begriff:


eine Stammfunktion F zur Funktion f

eine Stammfunktion F ist die Funktion, für die gilt:


F′(x)=f(x)F'(x)=f(x)


Definiere den Begriff:


Unbestimmtes Integral einer Funtktion

Das UNBESTIMMTE INTEGRAL:


∫f(x)dx\int f(x) dx


ist die Menge aller Stammfunktionen der Funktion. Es gilt:


∫f(x)dx=F(x)+c\int f(x) dx = F(x) + c


wobei:


c∈Rc \in \mathbb{R}


Das unbestimmte Integral ist also eine Funktionsschar.

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∫abf(x)dx\int_{a}^{b}f(x)dx


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∫abf(x)dx=[F(x)]ab=F(b)−F(a)\int_{a}^b f(x)dx=[F(x)]_{a}^b=F(b)-F(a)


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∫xndx\int x^n dx


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n≠0n \neq 0


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∫xndx=1n+1⋅xn+1+c‾‾\underline{\underline{\int x^n dx = \frac{1}{n+1}\cdot x^{n+1} + c}}


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c∈Rc \in \mathbb{R}


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∫exdx\int e^x dx





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∫exdx=ex+c‾‾\underline{\underline{\int e^x dx = e^x + c}}


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c∈Rc \in \mathbb{R}


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∫1xdx\int \frac{1}{x} dx





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∫1xdx=ln∣x∣+c‾‾\underline{\underline{\int \frac{1}{x} dx = ln|x| + c}}


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c∈Rc \in \mathbb{R}


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∫f(mx+n)dx\int f(mx+n)dx


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∫f(mx+n)=1m⋅F(mx+n)+c‾‾\underline{\underline{\int f(mx+n)=\frac{1}{m}\cdot F(mx+n)+c}}


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c∈Rc \in \mathbb{R}


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∫(g(x)+h(x))dx\int (g(x) + h(x)) dx


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∫(g(x)+h(x)dx=∫g(x)dx+∫h(x)dx+c‾‾\underline{\underline{\int (g(x)+h(x) dx = \int g(x) dx + \int h(x)dx +c}}


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c∈Rc \in \mathbb{R}


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∫(i(x)−k(x))dx\int (i(x)-k(x)) dx


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∫(i(x)−k(x))dx=∫i(x)dx−∫k(x)dx+c‾‾\underline{\underline{\int (i(x)-k(x)) dx=\int i(x)dx - \int k(x)dx+c}}


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c∈Rc \in \mathbb{R}


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∫(u(x)⋅v′(x))dx\int (u(x)\cdot v'(x)) dx


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∫(u(x)⋅v′(x))dx=u(x)⋅v(x)−∫(u′(x)⋅v(x))dx‾‾\underline{\underline{\int (u(x)\cdot v'(x))dx=u(x)\cdot v(x) - \int (u'(x)\cdot v(x)) dx}}


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A=∫abf(x)dx=F(b)−F(a)‾‾\underline{\underline{A=\int_{a}^bf(x)dx=F(b)-F(a)}}



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A=∣∫abf(x)dx∣=∣F(b)−F(a)∣‾‾\underline{\underline{A=|\int_{a}^b f(x)dx|=|F(b)-F(a)|}}

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∫xdx=\int x dx=


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∫xdx=12x2+c‾‾\underline{\underline{\int x dx = \frac{1}{2}x^2 + c}}


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c∈Rc \in \mathbb{R}


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∫x3dx=\int x^3 dx=


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∫x3dx=14x4+c‾‾\underline{\underline{\int x^3 dx = \frac{1}{4}x^4+c}}


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c∈Rc \in \mathbb{R}


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∫1x2dx=\int \frac{1}{x^2}dx=


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∫1x2=−1x1+c‾‾\underline{\underline{\int \frac{1}{x^2}=-\frac{1}{x^1} +c}}

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c∈Rc\in \mathbb{R}


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∫1dx=\int 1dx=


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∫1dx=x+c‾‾\underline{\underline{\int 1 dx = x+c}}


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c∈Rc \in \mathbb{R}


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∫(ex+1)dx=\int (e^x+1) dx=


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∫(ex+1)dx=ex+x+c‾‾\underline{\underline{\int (e^x+1)dx=e^x+x+c}}


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c∈Rc \in \mathbb{R}


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f(x)=exf(x)=e^x


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∫ln⁡(x)dx=\int \ln(x) dx=


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∫ln⁡(x)dx=x⋅ln⁡(x)−x+c‾‾\underline{\underline{\int \ln(x) dx=x\cdot \ln(x) - x +c}}


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c∈Rc \in \mathbb{R}


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g(x)=2xg(x)=2x


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∫e2x+3dx\int e^{2x+3}dx


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∫(4x−2)3dx\int (4x-2)^3 dx


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∫x2dx=13x3\int x^2dx=\frac{1}{3}x^3


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∫(x2−x)dx\int (x^2 -x)dx


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∫(x2−x)dx=13x3−12x2+c‾‾\underline{\underline{\int(x^2-x)dx=\frac{1}{3}x^3-\frac{1}{2}x^2+c}}


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∫141dx\int_{1}^{4}1dx


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∫141dx=[x]14=4−1=3‾‾\underline{\underline{\int_{1}^{4}1dx=[x]_{1}^{4}=4-1=3}}



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Author

Rick K.

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