$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

Somit gilt:

$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = 6 = 6

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

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

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

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

= $-\frac{2}{2}-\frac{1}{2}-1·\left(-10\right)$

= $-\frac{3}{2}+10$

= $-\frac{3}{2}+\frac{20}{2}$

= $\frac{17}{2}$

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

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

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

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

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

= ${\left[-\frac{4}{9}{\left(3x-7\right)}^{\frac{3}{2}}\right]}_{\frac{23}{3}}^{\frac{32}{3}}$

= ${\left[-\frac{4}{9}{\left(\sqrt{3x-7}\right)}^3\right]}_{\frac{23}{3}}^{\frac{32}{3}}$

$=$ $-\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{32}{3}\right)-7}\right)}^3+\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{23}{3}\right)-7}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{32-7}\right)}^3+\frac{4}{9}{\left(\sqrt{23-7}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{25}\right)}^3+\frac{4}{9}{\left(\sqrt{16}\right)}^3$

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

= $-\frac{4}{9}\cdot 125+\frac{4}{9}\cdot 64$

= $-\frac{500}{9}+\frac{256}{9}$

= $-\frac{244}{9}$

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

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

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

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

= $2+\frac{9}{2}\cdot {\pi }^3-\left(-2+\frac{1}{6}\cdot {\pi }^3\right)$

= $2+\frac{9}{2}\cdot {\pi }^3-1·\left(-2\right)-1·\frac{1}{6}\cdot {\pi }^3$

= $2+\frac{9}{2}\cdot {\pi }^3+2-\frac{1}{6}\cdot {\pi }^3$

= $2+2+\frac{9}{2}\cdot {\pi }^3-\frac{1}{6}\cdot {\pi }^3$

= $4+\frac{13}{3}\cdot {\pi }^3$

= ${\left[\frac{1}{3}{e}^{3x-3}\right]}_1^4$

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

= $\frac{1}{3}{e}^{12-3}-\frac{1}{3}{e}^{3-3}$

= $\frac{1}{3}{e}^9-\frac{1}{3}{e}^0$

= $\frac{1}{3}{e}^9-\frac{1}{3}$