$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

Somit gilt:

$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -3 +3 = 0

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

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

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

= $-\frac{32}{3}+4-\left(\frac{32}{3}+4\right)$

= $-\frac{32}{3}+\frac{12}{3}-\left(\frac{32}{3}+\frac{12}{3}\right)$

= $-\frac{20}{3}-1·\frac{44}{3}$

= $-\frac{20}{3}-\frac{44}{3}$

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

= ${\left[-\frac{1}{4}{x}^3-\frac{3}{4}{x}^{-3}\right]}_2^6$

= ${\left[-\frac{1}{4}{x}^3-\frac{3}{4x^3}\right]}_2^6$

$=$ $-\frac{1}{4}\cdot 6^3-\frac{3}{4\cdot 6^3}-\left(-\frac{1}{4}\cdot 2^3-\frac{3}{4\cdot 2^3}\right)$

= $-\frac{1}{4}\cdot 216-\frac{3}{4}\cdot \left(\frac{1}{216}\right)-\left(-\frac{1}{4}\cdot 8-\frac{3}{4}\cdot \left(\frac{1}{8}\right)\right)$

= $-54-\frac{1}{288}-\left(-2-\frac{3}{32}\right)$

= $-\frac{15552}{288}-\frac{1}{288}-\left(-\frac{64}{32}-\frac{3}{32}\right)$

= $-\frac{15553}{288}-1·\left(-\frac{67}{32}\right)$

= $-\frac{15553}{288}+\frac{67}{32}$

= $-\frac{15553}{288}+\frac{603}{288}$

= $-\frac{7475}{144}$

= ${\left[-\frac{2}{9}{\left(3x-4\right)}^{-3}\right]}_2^4$

= ${\left[-\frac{2}{9{\left(3x-4\right)}^3}\right]}_2^4$

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

= $-\frac{2}{9{\left(12-4\right)}^3}+\frac{2}{9{\left(6-4\right)}^3}$

= $-\frac{2}{9\cdot 8^3}+\frac{2}{9\cdot 2^3}$

= $-\frac{2}{9}\cdot \left(\frac{1}{512}\right)+\frac{2}{9}\cdot \left(\frac{1}{8}\right)$

= $-\frac{1}{2304}+\frac{1}{36}$

= $-\frac{1}{2304}+\frac{64}{2304}$

= $\frac{7}{256}$

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

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

= $-3\cdot 0-\left(-1\right)-\left(-3\cdot 0-1\right)$

= $0+1-\left(0-1\right)$

= $1+1$

= $2$

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

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

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

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

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

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

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

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

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

= $-\frac{37}{3}$