Aufgabe:
Doppelintegrale bilden.
Problem/Ansatz:
Ich bräuchte nur eine Überprüfung der Ergebnisse. 
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(1) a) \( \int \limits_{0}^{\frac{9}{4}} \int \limits_{x^{3}}^{\frac{9}{2}} x^{7} y d y d x \)
\( \begin{array}{l} \text { inneres: } \int \limits_{x^{9}}^{x^{\frac{9}{2}}} x^{7} y d y=x^{7} \cdot \int \limits_{x^{9}}^{x^{\frac{9}{2}}} y d y \cdot x^{7}\left[\frac{1}{2} y^{2}\right]_{x^{9}}^{x^{\frac{9}{2}}}=x^{7}\left[\frac{y^{2}}{2}\right]_{x^{9}}^{x^{\frac{9}{2}}} \\ =x^{7}\left(\frac{\left(x^{2}\right)^{2}}{2}-\frac{\left(x^{9}\right)^{2}}{2}\right) \\ =x^{7}\left(\frac{x^{9}}{2}-\frac{x^{8}}{2}\right) \\ \text { - } \frac{x^{15}}{2}-\frac{x^{25}}{2} \\ =\frac{x^{16}-x^{25}}{2} \\ \text { auberes } \int \limits_{0}^{\frac{9}{4}} \frac{x^{16}-x^{25}}{2} d x=\frac{1}{2} \int \limits_{0}^{\frac{9}{4}} x^{16}-x^{25} d x=\frac{1}{2}\left[\frac{1}{17} x^{17}-\frac{1}{26} x^{26}\right] \\ =\frac{1}{2}\left(\left(\frac{1}{17} \cdot\left(\frac{9}{4}\right)^{17}-\frac{1}{26}\left(\frac{9}{a}\right)^{26}\right)-\left(\frac{1}{17} \cdot 0^{17}-\frac{1}{26} \cdot 0^{26}\right)\right) \\ =-27560840,67 \\ \end{array} \)
b) \( \int \limits_{0}^{14} \int \limits_{0}^{-\frac{1}{2} x+\pi} x \cos (y) d y d x= \)
inneres: \( \int \limits_{0}^{-\frac{1}{3} x+\pi} x \cdot \cos (\gamma) d y=x \int \limits_{0}^{-\frac{1}{3} x+\pi} \cos (\gamma) d y=x[\sin (\gamma)]_{0}^{-\frac{1}{3} x+\pi} \)
\( \begin{array}{l} =x\left(\sin \left(-\frac{1}{8} x+\pi\right)-\sin (01)\right. \\ =x \cdot \sin \left(-\frac{1}{8} x+\pi\right) \\ \sin \left(-\frac{1}{8} x+\pi\right) \end{array} \)
auBeres: \( \int \limits_{0}^{14} x \cdot \sin \left(-\frac{1}{8} x+\pi\right) d x \)
partielle integration: \( U(x) \cdot v(x)-\int U^{\prime}(x) \cdot v(x) d x \)
\( \begin{array}{l} u(x)=x \quad v^{\prime}(x)=\sin \left(-\frac{1}{8} x+\pi\right) \\ u^{\prime}(x)=1 \quad v(x) \div(-9) \cos \left(-\frac{1}{8} x+\pi\right) \\ x \cdot(-9) \cdot \cos \left(-\frac{1}{9} x+\pi\right)-\int 1 \cdot(-9) \cos \left(-\frac{1}{9} x+\pi\right) d x \\ =-9 x \cdot \cos \left(-\frac{1}{8} x+\pi\right)-(-9) \cdot \int \limits_{0}^{14} \cos \left(-\frac{1}{9} x+\pi\right) d x \\ =\left[-9 x \cdot \cos \left(-\frac{1}{8} x+\pi\right)-(-9) \cdot(-9) \sin \left(-\frac{1}{8} x+\pi\right)\right]_{0}^{14} \\ =\left[-9 x \cdot \cos \left(-\frac{1}{9} x+\pi\right)-81 \cdot \sin \left(-\frac{1}{9} x+\pi\right)\right]_{0}^{14} \\ =\left[\left(\left(-9 \cdot 14 \cdot \cos \left(-\frac{1}{8} \cdot 11+\pi\right)-81 \cdot \sin \left(-\frac{1}{9} \cdot 14+\pi\right)\right)-\left(-81 \sin \left(-\frac{1}{8} \cdot 0+\pi\right)\right)\right.\right. \\ =-126 \cos \left(-\frac{14}{9}+\pi\right)-81 \sin \left(-\frac{14}{9}+\pi\right) \\ =79,07032896 \end{array} \)
