$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 6) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 -8 = -14

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

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

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

= $-\frac{2}{3}+4-\left(-\frac{16}{3}+8\right)$

= $-\frac{2}{3}+\frac{12}{3}-\left(-\frac{16}{3}+\frac{24}{3}\right)$

= $\frac{10}{3}-1·\frac{8}{3}$

= $\frac{10}{3}-\frac{8}{3}$

= $\frac{2}{3}$

= ${\left[-\cos \left(x\right)+\frac{3}{10}{x}^5\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\cos \left(\pi \right)+\frac{3}{10}\cdot {\pi }^5-\left(-\cos \left(\frac{1}{2}\pi \right)+\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $-\left(-1\right)+\frac{3}{10}\cdot {\pi }^5-\left(-0+\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $1+\frac{3}{10}\cdot {\pi }^5-\left(0+\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $1+\frac{3}{10}\cdot {\pi }^5-\frac{3}{320}\cdot {\pi }^5$

= $1+\frac{93}{320}\cdot {\pi }^5$

= ${\left[-\frac{2}{9}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

= ${\left[-\frac{2}{9}{\left(\sqrt{3x-6}\right)}^3\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

$=$ $-\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{31}{3}\right)-6}\right)}^3+\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3$

= $-\frac{2}{9}{\left(\sqrt{31-6}\right)}^3+\frac{2}{9}{\left(\sqrt{22-6}\right)}^3$

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

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

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

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

= $-\frac{122}{9}$

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

$=$ $-\frac{1}{4}\cdot \cos \left(\pi \right)+\frac{1}{4}\cdot \sin \left(\pi \right)-\left(-\frac{1}{4}\cdot \cos \left(0\right)+\frac{1}{4}\cdot \sin \left(0\right)\right)$

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

= $\frac{1}{4}+0-\left(-\frac{1}{4}+0\right)$

= $\frac{1}{4}+0-\left(-\frac{1}{4}+0\right)$

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

= $\frac{1}{2}$

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

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

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

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

= $\frac{1}{2}\cdot \left(-8\right)+5-\left(\frac{1}{2}\cdot \left(-64\right)+0\right)$

= $-4+5-\left(-32+0\right)$

= $1+32$

= $33$