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

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

Somit gilt:

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

= ${\left[-\frac{1}{3}{x}^3+\frac{5}{2}{x}^2\right]}_{-2}^{-1}$

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

= $-\frac{1}{3}\cdot \left(-1\right)+\frac{5}{2}\cdot 1-\left(-\frac{1}{3}\cdot \left(-8\right)+\frac{5}{2}\cdot 4\right)$

= $\frac{1}{3}+\frac{5}{2}-\left(\frac{8}{3}+10\right)$

= $\frac{2}{6}+\frac{15}{6}-\left(\frac{8}{3}+\frac{30}{3}\right)$

= $\frac{17}{6}-1·\frac{38}{3}$

= $\frac{17}{6}-\frac{38}{3}$

= $\frac{17}{6}-\frac{76}{6}$

= $-\frac{59}{6}$

= ${\left[-\frac{4}{3}{x}^3-\frac{5}{2}\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

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

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

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

= $-\frac{1}{6}\cdot {\pi }^3-\left(0-\frac{5}{2}\right)$

= $-\frac{1}{6}\cdot {\pi }^3+\frac{5}{2}$

= ${\left[\frac{1}{2}{\left(-2x+2\right)}^3\right]}_1^2$

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

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

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

= $\frac{1}{2}\cdot \left(-8\right)-\frac{1}{2}\cdot 0$

= $-4+0$

= $-4$

= ${\left[-\frac{3}{5}{x}^5+4x^{\frac{5}{4}}\right]}_4^9$

= ${\left[-\frac{3}{5}{x}^5+4{\left(\sqrt[4]{x}\right)}^5\right]}_4^9$

$=$ $-\frac{3}{5}\cdot 9^5+4{\left(\sqrt[4]{9}\right)}^5-\left(-\frac{3}{5}\cdot 4^5+4{\left(\sqrt[4]{4}\right)}^5\right)$

= $-\frac{3}{5}\cdot 59049+4\cdot 1^5-\left(-\frac{3}{5}\cdot 1024+4\cdot 1^5\right)$

= $-\frac{177147}{5}+4\cdot 1-\left(-\frac{3072}{5}+4\cdot 1\right)$

= $-\frac{177147}{5}+4-\left(-\frac{3072}{5}+4\right)$

= $-\frac{177147}{5}+\frac{20}{5}-\left(-\frac{3072}{5}+\frac{20}{5}\right)$

= $-\frac{177127}{5}-1·\left(-\frac{3052}{5}\right)$

= $-\frac{177127}{5}+\frac{3052}{5}$

= $-34815$

= ${\left[-\frac{1}{3}{\left(-2x+5\right)}^{\frac{3}{2}}\right]}_{-\frac{11}{2}}^{-2}$

= ${\left[-\frac{1}{3}{\left(\sqrt{-2x+5}\right)}^3\right]}_{-\frac{11}{2}}^{-2}$

$=$ $-\frac{1}{3}{\left(\sqrt{-2\cdot \left(-2\right)+5}\right)}^3+\frac{1}{3}{\left(\sqrt{-2\cdot \left(-\frac{11}{2}\right)+5}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{4+5}\right)}^3+\frac{1}{3}{\left(\sqrt{11+5}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{9}\right)}^3+\frac{1}{3}{\left(\sqrt{16}\right)}^3$

= $-\frac{1}{3}\cdot 3^3+\frac{1}{3}\cdot 4^3$

= $-\frac{1}{3}\cdot 27+\frac{1}{3}\cdot 64$

= $-9+\frac{64}{3}$

= $-\frac{27}{3}+\frac{64}{3}$

= $\frac{37}{3}$