$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ gibt den orientierten (also mit Vorzeichen behafteten) Flächeninhalt zwischen dem Graph der Funktion f und der x-Achse an.

Um diesen aus dem Schaubild zu berechnen unterteilen wir die Fläche in Rechtecke, Parallelogramme und ggf. Dreiecke:

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

Somit gilt:

$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = 4.5 = 4.5

= ${\left[-\frac{5}{2}{x}^2+5x\right]}_0^2$

$=$ $-\frac{5}{2}\cdot 2^2+5\cdot 2-\left(-\frac{5}{2}\cdot 0^2+5\cdot 0\right)$

= $-\frac{5}{2}\cdot 4+10-\left(-\frac{5}{2}\cdot 0+0\right)$

= $-10+10-\left(0+0\right)$

= $0+0$

= $0$

= ${\left[9\cdot \sin \left(x\right)-\frac{5}{3}{e}^{3x}\right]}_0^{\frac{1}{2}\pi }$

$=$ $9\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(9\cdot \sin \left(0\right)-\frac{5}{3}{e}^{3\cdot \left(0\right)}\right)$

= $9\cdot 1-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(9\cdot 0-\frac{5}{3}{e}^0\right)$

= $9-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(0-\frac{5}{3}\right)$

= $-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}+9-\left(0-\frac{5}{3}\right)$

= $-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}+9+\frac{5}{3}$

= $-\frac{5}{3}{e}^{\frac{3}{2}\pi }+\frac{32}{3}$

= ${\left[-\frac{2}{9}{\left(-3x+4\right)}^{\frac{3}{2}}\right]}_{-7}^{-4}$

= ${\left[-\frac{2}{9}{\left(\sqrt{-3x+4}\right)}^3\right]}_{-7}^{-4}$

$=$ $-\frac{2}{9}{\left(\sqrt{-3\cdot \left(-4\right)+4}\right)}^3+\frac{2}{9}{\left(\sqrt{-3\cdot \left(-7\right)+4}\right)}^3$

= $-\frac{2}{9}{\left(\sqrt{12+4}\right)}^3+\frac{2}{9}{\left(\sqrt{21+4}\right)}^3$

= $-\frac{2}{9}{\left(\sqrt{16}\right)}^3+\frac{2}{9}{\left(\sqrt{25}\right)}^3$

= $-\frac{2}{9}\cdot 4^3+\frac{2}{9}\cdot 5^3$

= $-\frac{2}{9}\cdot 64+\frac{2}{9}\cdot 125$

= $-\frac{128}{9}+\frac{250}{9}$

= $\frac{122}{9}$

= ${\left[-\frac{1}{15}{x}^5-3\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{1}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5-3\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{1}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5-3\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{1}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5-3\cdot \left(-1\right)-\left(-\frac{1}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5-3\cdot 1\right)$

= $-\frac{1}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5+3-\left(-\frac{1}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5-3\right)$

= $3-\frac{81}{160}\cdot {\pi }^5-\left(-3-\frac{1}{480}\cdot {\pi }^5\right)$

= $3-\frac{81}{160}\cdot {\pi }^5-1·\left(-3\right)-1·\left(-\frac{1}{480}\cdot {\pi }^5\right)$

= $3-\frac{81}{160}\cdot {\pi }^5+3+\frac{1}{480}\cdot {\pi }^5$

= $3+3-\frac{81}{160}\cdot {\pi }^5+\frac{1}{480}\cdot {\pi }^5$

= $6-\frac{121}{240}\cdot {\pi }^5$

= ${\left[-{\left(-2x+3\right)}^{\frac{3}{2}}\right]}_{-11}^{-\frac{13}{2}}$

= ${\left[-{\left(\sqrt{-2x+3}\right)}^3\right]}_{-11}^{-\frac{13}{2}}$

$=$ $-{\left(\sqrt{-2\cdot \left(-\frac{13}{2}\right)+3}\right)}^3+{\left(\sqrt{-2\cdot \left(-11\right)+3}\right)}^3$

= $-{\left(\sqrt{13+3}\right)}^3+{\left(\sqrt{22+3}\right)}^3$

= $-{\left(\sqrt{16}\right)}^3+{\left(\sqrt{25}\right)}^3$

= $-4^3+5^3$

= $-64+125$

= $61$