Areal mellom graf og x-aksen :

dersom arealet ligger under a-aksen blir det negativt.

$a , b , c$ er nullpunktene til grafen :

\begin{align*}A&=\bigg|\int_a^b f(x) \ dx\bigg|+\bigg|\int_b^c f(x) \ dx\bigg|\end{align*}

Oppgave 1

\begin{align}\int \limits_0^4 (x+2) \ dx=\end{align}

  • \begin{align*}\int\limits_0^4 x+2 \ dx&=\bigg[\frac{1}{2}x^2+2x\bigg]_0^4\\&=\bigg(\frac{1}{2}\cdot 4^2+2\cdot 4\bigg)-0\\&=8+8=16\end{align*}

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Oppgave 2

\begin{align}\int \limits_{-2}^2 x^2 \ dx=\end{align}

  • \begin{align*}\int\limits_{-2}^2 x^2 \ dx&=\bigg[\frac{1}{3}x^3\bigg]_{-2}^2\\&=\frac{1}{3}(2^3-(-2)^3)\\&=\frac{1}{3}(8+8)\\&=\frac{16}{3}\end{align*}

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Oppgave 3

\begin{align}\int \limits_1^e \frac{1}{x} \ dx=\end{align}

  • \begin{align*}\int\limits_1^e \frac{1}{x} \ dx&=\big[\ln x\big]_1^e\\&=\ln e-\ln 1\\&=1-0\\&=1\end{align*}

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Oppgave 4

\begin{align}\int \limits_1^4 (\frac{1}{4} x^2-x+4) \ dx=\end{align}

  • \begin{align*}\int\limits_1^4 \frac{1}{4}x^2-x+4 \ dx&=\bigg[\frac{1}{12}x^3-\frac{1}{2}x^2+4x\bigg]_1^4\\&=\bigg(\frac{1}{12}\cdot 4^3-\frac{1}{2}\cdot 4^2+4\cdot 4\bigg)-\bigg(\frac{1}{12}\cdot 1^3-\frac{1}{2}\cdot 1^2+4\cdot 1\bigg)\\&=4^2\bigg(\frac{4}{12}-\frac{1}{2}+1\bigg)-\bigg(\frac{1}{12}-\frac{6}{12}+\frac{48}{12}\bigg)\\&=16\cdot \frac{5}{6}-\frac{43}{12}\\&=\frac{160-43}{12}\\&=\frac{117}{12}=\frac{39}{4}\end{align*}

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Oppgave 5

\begin{align}\int \limits_0^{\ln2} e^{2x} \ dx=\end{align}

  • \begin{align*}\int\limits_0^{\ln 2} e^{2x} \ dx&=\bigg[\frac{1}{2}e^{2x}\bigg]_0^{\ln 2}\\&=\frac{1}{2}(e^{2\ln 2}-e^0)\\&=\frac{1}{2}(4-1)\\&=\frac{3}{2}\end{align*}

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Oppgave 6

\begin{align}\int\limits_{-3}^2 -x^2-x+2 \ dx=\end{align}

  • \begin{align*}\int\limits_{-3}^2 -x^2-x+2 \ dx&=\bigg[-\frac{1}{3}x^3-\frac{1}{2}x^2+2x\bigg]_{-3}^2\\&=\bigg(-\frac{1}{3}\cdot 2^3-\frac{1}{2}\cdot 2^2+2\cdot 2 \bigg)-\bigg(-\frac{1}{3}\cdot (-3)^3-\frac{1}{2}\cdot (-3)^2+2\cdot (-3)\bigg)\\&=2^2\bigg(-\frac{2}{3}-\frac{1}{2}+1\bigg)-(-3)\bigg(-3+\frac{3}{2}+2\bigg)\\&=4\bigg(\frac{-4-3+6}{6}\bigg)+3\bigg(\frac{3-2}{2}\bigg)\\&=\frac{4\cdot (-1)}{6}+\frac{3}{2}\\&=\frac{-4+9}{6}\\&=\frac{5}{6}\end{align*}

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Oppgave 7

\begin{align}\int\limits_0^2 \frac{1}{3}x^2-3 \ dx=\end{align}

  • \begin{align*}\int\limits_0^2 \frac{1}{3}x^2-3 \ dx&=\bigg[\frac{1}{3}\cdot \frac{1}{3}x^3-3x\bigg]_0^2\\&=\bigg(\frac{1}{3}\cdot \frac{1}{3}\cdot 2^3-3\cdot 2\bigg)-0\\&=\frac{8}{9}-6\\&=\frac{8-54}{9}=\frac{-46}{9}\end{align*}

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Oppgave 8

\begin{align}\int\limits_0^2 e^x \ dx=\end{align}

  • \begin{align*}\int\limits_0^2 e^x \ dx&=[e^x]_0^2\\&=e^2-e^0\\&=e^2-1\end{align*}

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Oppgave 9

\begin{align}\int\limits_0^2 \sqrt{x} \ dx=\end{align}

  • \begin{align*}\int\limits_0^2 \sqrt{x} \ dx&=\int\limits_0^2 x^{\frac{1}{2}} \ dx\\&=\bigg[\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}\bigg]_0^2\\&=\bigg[ \frac{2}{3} x^{\frac{3}{2}}\bigg]_0^2\\&=\bigg[ \frac{2}{3} x\sqrt{x}\bigg]_0^2\\&=\bigg( \frac{2}{3} \cdot 2\cdot \sqrt{2}\bigg)-0\\&=\frac{4\sqrt{2}}{3}\end{align*}

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Oppgave 10

\begin{align}\int \limits_0^2 \frac{2x^2}{x^3+1} \ dx=\end{align}

  •  \begin{align*}\int \frac{2x^2}{x^3+1} \ dx&=2\int \frac{x^2}{u}\cdot \frac{1}{3x^2} \ du\\u&= x^3+1 , u'=3x^2\\&=\frac{2}{3}\int \frac{1}{u} \ du\\&=\frac{2}{3}\cdot \ln|u|+C\\&=\frac{2}{3}\cdot \ln|x^2+1|+C\\\int\limits_0^2 \frac{2x^2}{x^3+1} \ dx&=\Big[\frac{2}{3}\cdot \ln|x^3+1|\Big]_0^2\\&=\frac{2}{3}(\ln|2^3+1|-\ln|0+1|)\\&=\frac{2}{3}(\ln|3^2|-\ln|1|)\\&=\frac{4\ln 3}{3}\\\end{align*}

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  • \begin{align*}\int x\cdot\ln x \ dx&=\frac{1}{2}x^2\ln x-\int \frac{1}{2}x^2\cdot \frac{1}{x} \ dx\\&=\frac{1}{2}x^2\ln x-\frac{1}{2}\cdot \frac{1}{2} x^2+C\\\int\limits_0^e x\cdot\ln x \ dx&=\frac{1}{2}\bigg[x^2\ln x- \frac{1}{2} x^2\bigg]_0^e\\&=\frac{1}{2}\bigg(e^2-\frac{1}{2}e^2\bigg)-0\\&=\frac{e^2}{4}\end{align*}

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Oppgave 11

\begin{align}\int \limits_0^e x\cdot \ln x \ dx=\end{align}

Oppgave 12

\begin{align}\int \limits_1^e \frac{3}{x} \ dx=\end{align}

  •  \begin{align*}\int\limits_1^e \frac{3}{x} \ dx&=3\int\limits_1^e \frac{1}{x} \ dx\\&=3\bigg[\ln x\bigg]_1^e\\&=3(\ln e-\ln 1)\\&=3\end{align*}

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Oppgave 13

\begin{align}\int\limits_0^4 x+2 \ dx\end{align}

  •  \begin{align*}\int\limits_0^4 x+2 \ dx&=\bigg[\frac{1}{2}x^2+2x\bigg]_0^4\\&=\bigg(\frac{1}{2}\cdot 4^2+2\cdot 4\bigg)-0\\&=8+8=16\end{align*}

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  • \begin{align*}f(x)&=0\\x-2&=0\\x&=2\\\text{vi må dele opp integralet}&\text{ fordi det er et nullpunkt i intervallet}\\\int x-2 \ dx&=\frac{1}{2}x^2-2x+C\\\int_0^2 x-2 \ dx&=\bigg[\frac{1}{2}x^2-2x\bigg]_0^2\\&=(\frac{1}{2}\cdot 2^2-2\cdot 2)-0\\&=-2\\\int_2^4 x-2 \ dx&=\bigg[\frac{1}{2}x^2-2x\bigg]_2^4\\&=(\frac{1}{2}\cdot 4^2-2\cdot 4)-(\frac{1}{2}\cdot 2^2-2\cdot 2)\\&=8-8-(-2)\\&=2\\A &=2+2=4\end{align*}

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Oppgave 14

Finn arealet som er avgrenset av definisjonsområdet.\begin{align}f(x)=x-2 , x\in[0,4]\end{align}

Oppgave 15

Finn arealet som er avgrenset av definisjonsområdet. \begin{align}f(x)=x^2-4, x\in[0,4]\end{align}

  • \begin{align*}f(x)&=x^2-4\\f(x)&=0\\x&=\pm 2\\\int_0^2 x^2-4 \ dx&=\bigg[ \frac{1}{3}x^3-4x\bigg]_0^2\\&=( \frac{1}{3}\cdot 2^3-4\cdot 2)-0\\&=\frac{8}{3}-8=\frac{8-24}{3}=\frac{-16}{3}\\\int_2^4 x^2-4 \ dx&=\bigg[ \frac{1}{3}x^3-4x\bigg]_2^4\\&=( \frac{1}{3}\cdot 4^3-4\cdot 4)-( \frac{1}{3}\cdot 2^3-4\cdot 2)\\&=(\frac{64}{3}-16)-(\frac{8}{3}-8)\\&=\frac{64-48}{3}-\frac{8-24}{3}=\frac{-16}{3}\\&=\frac{16-(-16)}{3}=\frac{32}{3}\\A &=\frac{16}{3}+\frac{32}{3}=\frac{48}{3}\end{align*}

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Oppgave 16

Vi har gitt funksjonen

\begin{align}f(x)=-\frac{1}{2}x^2+2 ,x\in[-2,2] \end{align}

  • \begin{align*}f(x)&=-\frac{1}{2}x^2+2 \ dx\\f(x)&=0\\x&=\pm 2\\\int_{-2}^2 -\frac{1}{2}x^2+2 \ dx&=\bigg[-\frac{1}{6}x^3+2x \bigg]_{-2}^2\\&=\bigg(-\frac{1}{6}\cdot 2^3+2\cdot 2 \bigg)-\bigg(-\frac{1}{6}\cdot (-2)^3+2\cdot (-2) \bigg)\\&=-\frac{4}{3}+4-\frac{4}{3}+4\\&=-\frac{8}{3}+\frac{26}{3}\\A &=\frac{16}{3}\end{align*}

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Oppgave 17

Finn arealet som er avgrenset av grafene til $f(x)=\frac{1}{3}x^3-3$ og grafen til $g(x)=x-3$.

  • \begin{align*}f(x)&=g(x)\\\frac{1}{3}x^3-3&=x-3\\\frac{1}{3}x^3-x&=0\\\frac{1}{3}x(x^2-3)&=0\\x&\in\{ -\sqrt{3} , 0 , \sqrt{3}\}\end{align*}

    \begin{align*}A_1 &= \int\limits_{-\sqrt{3}}^0 \frac{1}{3}x^3-x \ dx \\&= \bigg[\frac{1}{3}\cdot\frac{1}{4}x^4-\frac{1}{2}x^2\bigg]_{-\sqrt{3}}^0 \\&=0-\bigg(\frac{(-\sqrt{3})^4}{12}-\frac{1}{2}\cdot (-\sqrt{3})^2\bigg)\\&=-(\frac{9}{12}-\frac{3}{2})=-\frac{3}{4}+\frac{6}{4}=-\frac{3}{4}\end{align*}

    \begin{align*}A_2 &= \int\limits_0^{\sqrt{3}} \frac{1}{3}x^3-x \ dx\\&= \bigg[\frac{1}{3}\cdot\frac{1}{4}x^4-\frac{1}{2}x^2\bigg]_0^{\sqrt{3}} \\&=\bigg(\frac{(\sqrt{3})^4}{12} -\frac{\sqrt{3}^2}{2} \bigg)-0\\&=\frac{9}{12}-\frac{3}{2}=\frac{9}{12}-\frac{18}{12}=-\frac{9}{12}=-\frac{3}{4}\end{align*}

    $$A=|A_1|+|A_2|=\frac{3}{4}+\frac{3}{4}=\frac{6}{4}=\frac{3}{2}$$

Oppgave 18

Vi har gitt 2 funksjoner, $f(x)=x^2-2x+1$ og $g(x)=a\cdot x+1$\\

Hvilken verdi må $a$ ha for at arealet mellom f og g skal være null.

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Oppgave 19

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