Das ist nun mein Ergebnis Stimmt das so?
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\( \begin{array}{l} 1+x=(2 x)^{2} \\ 1+x-4 x^{2} \quad 1-4 x^{2} \\ -4 x^{2}+x+1=0 \quad 1 \cdot(-4) \\ x^{2}-\frac{1}{4 x}-\frac{1}{4}=0 \quad 1 p / q \\ x_{1}=\frac{1+\sqrt{17}}{8} \\ x_{2}=-0,390388 \\ \end{array} \)
Greme
\( \int \limits_{0}^{\frac{1+\sqrt{1}}{8}} \int \limits_{(2 x)^{2}}^{1+x} \sqrt{x} y d y d x \)
inneres.
\( \begin{aligned} \int \limits_{4 x^{2}}^{1+x} \sqrt{x} y d y=\sqrt{x}\left[\frac{1}{2} y^{2}\right]_{4 x^{2}}^{1+x} & =\sqrt{x} \cdot \frac{1}{2}\left((1+x)^{2}-\left(4 x^{2}\right)^{2}\right) \\ & =\sqrt{x} \cdot \frac{1}{2}\left(x^{2}+2 x+1-16 x^{4}\right) \end{aligned} \)
\( \frac{1+\sqrt{17}}{8} \)
außeres: \( \int \limits_{0}^{8} \sqrt{x} \cdot \frac{1}{2} \cdot\left(x^{2}+2 x+1-16 x^{4}\right) d x \)
\( \begin{array}{l} =-8 \int x^{4,5} d x+\frac{1}{2} \int x^{2,5} d x+\int x^{1,5} d x+\frac{1}{2} \int \sqrt{x} d x \\ =-8\left[\frac{2}{11} x^{5,5}\right]+\frac{1}{2}\left[\frac{2}{7} x^{3,5}\right]+\left[\frac{2}{6} x^{215}\right]+\frac{1}{2}\left[\frac{2}{3} x^{\frac{1}{2}}\right] \\ =-\frac{16}{1}\left[x^{5,5}\right]+\frac{1}{7}\left[x^{3,5}\right]+\frac{2}{5}\left[x^{2,5}\right]+\frac{1}{3}\left[x^{\frac{3}{2}}\right] \\ =-\frac{16}{10}\left(\left(\frac{1+\sqrt{17}}{8}\right)^{3 / 5}-0^{3,5}\right)+\frac{1}{7}\left(\left(\frac{1+\sqrt{17}}{8}\right)^{3,5}-0^{3,5}\right)+\frac{2}{5}\left(\left(\frac{1+\sqrt{10}}{8}\right)^{2,5}-0^{2,5}\right)+\frac{1}{3}\left(\left(\frac{1+\sqrt{17}}{8}\right)^{\frac{1}{2}}-0^{\frac{3}{2}}\right) \\ \text { - } 0,035931901375+\frac{1}{3}\left(\left(\frac{1+a_{4}}{8}\right)^{\frac{3}{2}}-0^{\frac{3}{2}}\right) \\ \end{array} \)