Matematik

Bestemte integraler

16. januar 2020 af Larsdk4 (Slettet) - Niveau: A-niveau

En som kan hjælpe mig videre

Vedhæftet fil: Udklip.PNG

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Svar #1
16. januar 2020 af Mathias7878

Hvis du gerne have mellemregninger med, som du kan forstå, så kan du bare bruge symbolab.com. Den beskriver de forskellige trin.

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Svar #2
16. januar 2020 af Larsdk4 (Slettet)

#1
Hvis du gerne have mellemregninger med, som du kan forstå, så kan du bare bruge symbolab.com. Den beskriver de forskellige trin.

Tak :)

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Svar #3
16. januar 2020 af mathon

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Svar #4
16. januar 2020 af mathon

$\small \begin{array}{lllll}b)&\int_{1}^{10}\frac{x^3+x}{x}\, \mathrm{d}x=\int_{1}^{10}(x^2+1)\, \mathrm{d}x=\left [ \frac{1}{3}x^3+x \right ]_{1}^{10}= \frac{1}{3}\cdot 10^3+10 -\left ( \frac{1}{3}\cdot 1^3+1 \right )=\\\\&\frac{1000}{3}+10-\frac{1}{3}-1=\frac{999}{3}+\frac{1}{3}+10-\frac{1}{3}-1=333+9=342 \end{array}$

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Svar #5
16. januar 2020 af mathon

$\small \small \begin{array}{lllll}c)&\int\left ( \ln(x)\cdot x^{-2} \right )\, \mathrm{d}x=- \ln(x)\cdot \frac{1}{x}-\int\left (\left ( -\frac{1}{x} \right )\cdot \frac{1}{x} \right ) \, \mathrm{d}x=- \ln(x)\cdot \frac{1}{x}+\int x^{-2}\mathrm{d}x=\\\\&- \ln(x)\cdot \frac{1}{x}-\frac{1}{x}=-\frac{1}{x}\left ( \ln(x)+1 \right )\\\\&\int_{0.2}^{0.4}\left ( \ln(x)\cdot x^{-2} \right )\, \mathrm{d}x=\left [ -\frac{1}{x}\left ( \ln(x)+1 \right ) \right ]_{0.2}^{0.4}= -\frac{1}{0.4}\left ( \ln(0.4)+1 \right ) -\left ( -\frac{1}{0.2}\left ( \ln(0.2)+1 \right ) \right )=\\\\&-3.25646 \end{array}$

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Svar #6
16. januar 2020 af mathon

$\small \small \begin{array}{lllll}d)&\int\left ( \ln(x)\cdot x^{2} \right )\, \mathrm{d}x=\left ( x\ln(x)-x \right )\cdot x^2-\int \left ( x\ln(x)-x \right )\cdot 2x\, \mathrm{d}x=\\\\& \left ( x\ln(x)-x \right )\cdot x^2-2\int \left ( x^2\ln(x)-x^2 \right )\, \mathrm{d}x=\\\\& \left ( x\ln(x)-x \right )\cdot x^2-2\int(\ln(x)\cdot x^2)\, \mathrm{d}x-2\int x^2\, \mathrm{d}x=\\\\&\int\left ( \ln(x)\cdot x^{2} \right )\, \mathrm{d}x=\left ( x\ln(x)-x \right )\cdot x^2-2\int(\ln(x)\cdot x^2)\, \mathrm{d}x-2\int x^2\, \mathrm{d}x\\\\&3\int\left ( \ln(x)\cdot x^{2} \right )\, \mathrm{d}x=\left ( x\ln(x)-x \right )\cdot x^2-\frac{2}{3}x^3\\\\&\int\left ( \ln(x)\cdot x^{2} \right )\, \mathrm{d}x=\frac{1}{3}x^2\left ( x\ln(x)-x \right )-\frac{2}{3}x^3\\\\\\&\int_{0.2}^{0.4}\left ( \ln(x)\cdot x^{2} \right )\, \mathrm{d}x=\left [\frac{1}{3}x^2\left ( x\ln(x)-x \right )-\frac{2}{3}x^3-\left ( \frac{1}{3}x^2\left ( x\ln(x)-x \right )-\frac{2}{3}x^3\right) \right ]_{0.2}^{0.4}=\\\\ \end{array}$

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Svar #7
16. januar 2020 af mathon

$\small \begin{array}{llllll}\textup{korrektion:}\\d)&\int (\ln(x)\cdot x^2)\, \mathrm{d}x=\left ( x\ln(x)-x \right )\cdot x^2-\int (x\ln(x)-x)\cdot 2x\, \mathrm{d}x= \\\\&\int (\ln(x)\cdot x^2)\, \mathrm{d}x=x^3\left ( \ln(x)-1 \right )-2\int \left (\ln(x)\cdot x^2-x^2 \right )\, \mathrm{d}x\\\\&\int (\ln(x)\cdot x^2)\, \mathrm{d}x=x^3\left ( \ln(x)-1 \right )-2\int \ln(x)\cdot x^2\, \mathrm{d}x+2\int x^2\, \mathrm{d}x\\\\&3\int (\ln(x)\cdot x^2)\, \mathrm{d}x=x^3\left ( \ln(x)-1 \right )+\frac{2}{3}x^3+k_1\\\\&\int (\ln(x)\cdot x^2)\, \mathrm{d}x=\frac{1}{3}x^3\left ( \ln(x)-1 \right )+\frac{2}{9}x^3+k\\\textup{og}\\\\& \int_{0.2}^{0.4} (\ln(x)\cdot x^2)\, \mathrm{d}x=\left [\frac{1}{3}x^3\left ( \ln(x)-1 \right )+\frac{2}{9}x^3 \right ]_{0.2}^{0.4}=-0.021478 \end{array}$

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Svar #8
17. januar 2020 af mathon

$\small \small \small \begin{array}{lllll}e)&\int (x\cdot 3^{-x})\, \mathrm{d}x=-\frac{1}{\ln(3)}\cdot 3^{-x}+\frac{1}{\ln(3)}\int 3^{-x}\, \mathrm{d}x=-\frac{1}{\ln(3)}\cdot 3^{-x}-\frac{1}{\ln^2(3)}\cdot 3^{-x}\\\\\\&\int_{1}^{120}x\cdot 3^{-x}\, \mathrm{d}x=-\frac{1}{\ln(3)}\cdot 3^{-120}-\frac{1}{\ln^2(3)}\cdot 3^{-120}-\left ( -\frac{1}{\ln(3)}\cdot 3^{-1}-\frac{1}{\ln^2(3)}\cdot 3^{-1} \right )=0.579592 \end{array}$

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