$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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

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

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

= $\frac{2}{3}\cdot 27+\frac{1}{2}\cdot 9-\left(\frac{2}{3}\cdot \left(-1\right)+\frac{1}{2}\cdot 1\right)$

= $18+\frac{9}{2}-\left(-\frac{2}{3}+\frac{1}{2}\right)$

= $\frac{36}{2}+\frac{9}{2}-\left(-\frac{4}{6}+\frac{3}{6}\right)$

= $\frac{45}{2}-1·\left(-\frac{1}{6}\right)$

= $\frac{45}{2}+\frac{1}{6}$

= $\frac{135}{6}+\frac{1}{6}$

= $\frac{68}{3}$

= ${\left[\frac{7}{3}\cdot \sin \left(x\right)-\frac{2}{9}{x}^3\right]}_0^{\pi }$

$=$ $\frac{7}{3}\cdot \sin \left(\pi \right)-\frac{2}{9}\cdot {\pi }^3-\left(\frac{7}{3}\cdot \sin \left(0\right)-\frac{2}{9}\cdot {\left(0\right)}^3\right)$

= $\frac{7}{3}\cdot 0-\frac{2}{9}\cdot {\pi }^3-\left(\frac{7}{3}\cdot 0-\frac{2}{9}\cdot 0\right)$

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

= $-\frac{2}{9}\cdot {\pi }^3+0$

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

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

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

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

= $\frac{1}{2}\cdot 0-\frac{1}{2}\cdot 0$

= $0+0$

= 0

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

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

= $-4\cdot \left(-1\right)+5\cdot 0-\left(-4\cdot 1+5\cdot 0\right)$

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

= $4+4$

= $8$

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

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

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

= $\frac{1}{3}\cdot 1-\frac{1}{3}\cdot 0$

= $\frac{1}{3}+0$

= $\frac{1}{3}+0$

= $\frac{1}{3}$