$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ 1 = 3 ⋅ 1 = 3.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 1.5 +3 = 4.5

= ${\left[-\frac{2}{3}{x}^3+5x\right]}_{-2}^2$

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

= $-\frac{2}{3}\cdot 8+10-\left(-\frac{2}{3}\cdot \left(-8\right)-10\right)$

= $-\frac{16}{3}+10-\left(\frac{16}{3}-10\right)$

= $-\frac{16}{3}+\frac{30}{3}-\left(\frac{16}{3}-\frac{30}{3}\right)$

= $\frac{14}{3}-1·\left(-\frac{14}{3}\right)$

= $\frac{14}{3}+\frac{14}{3}$

= $\frac{28}{3}$

= ${\left[4\cdot \cos \left(x\right)-\frac{5}{8}{e}^{-2x}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $4\cdot \cos \left(\frac{3}{2}\pi \right)-\frac{5}{8}{e}^{-2\cdot \left(\frac{3}{2}\pi \right)}-\left(4\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{5}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $4\cdot 0-\frac{5}{8}{e}^{-2\cdot \left(\frac{3}{2}\pi \right)}-\left(4\cdot 0-\frac{5}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $0-\frac{5}{8}{e}^{-2\cdot \left(\frac{3}{2}\pi \right)}-\left(0-\frac{5}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{5}{8}{e}^{-3\pi }+\frac{5}{8}{e}^{-\pi }$

= ${\left[-\frac{1}{4}{\left(x-3\right)}^4\right]}_0^2$

$=$ $-\frac{1}{4}\cdot {\left(2-3\right)}^4+\frac{1}{4}\cdot {\left(0-3\right)}^4$

= $-\frac{1}{4}\cdot {\left(-1\right)}^4+\frac{1}{4}\cdot {\left(-3\right)}^4$

= $-\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 81$

= $-\frac{1}{4}+\frac{81}{4}$

= $20$

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

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

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

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

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

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

= $\frac{5}{3}{e}^{\frac{3}{2}\pi }-\frac{11}{3}$

= ${\left[-\frac{2}{3}{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[-\frac{2}{3}{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $-\frac{2}{3}{\left(\sqrt{28-3}\right)}^3+\frac{2}{3}{\left(\sqrt{19-3}\right)}^3$

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

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

= $-\frac{2}{3}\cdot 125+\frac{2}{3}\cdot 64$

= $-\frac{250}{3}+\frac{128}{3}$

= $-\frac{122}{3}$