\begin{align}\int a^b f(x) \ dx =\Big[ F(x)\Big]_{a}^{b}= F(b)-F(a)\end{align}

Oppgave 1

Løs disse bestemte integralene for hånd

  • \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*}

  • \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*}

  • \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*}

  • \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*}

  • \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*}

Oppgave 2

  • \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*}

  • \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*}

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

Oppgave 3

  • \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*}

  • \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*}

  • \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*}

  • \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*}

Oppgave 4

  • \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*}

  • \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*}

  • \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*}