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

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

Somit gilt:

$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -4 -8 = -12

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

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

= $1-2-\left(9+6\right)$

= $1-2-1·9-1·6$

= $1-2-9-6$

= $-16$

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

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

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

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

= $-\frac{4}{3}\cdot 8+\frac{2}{3}\cdot \sin \left(4\right)-\left(-\frac{4}{3}\cdot 0+0\right)$

= $-\frac{32}{3}+\frac{2}{3}\cdot \sin \left(4\right)-\left(0+0\right)$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{32}{3}+0$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{32}{3}$

= ${\left[\frac{4}{9}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_3^{\frac{14}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-5}\right)}^3\right]}_3^{\frac{14}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{14}{3}\right)-5}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot 3-5}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{14-5}\right)}^3-\frac{4}{9}{\left(\sqrt{9-5}\right)}^3$

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

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

= $\frac{4}{9}\cdot 27-\frac{4}{9}\cdot 8$

= $12-\frac{32}{9}$

= $\frac{108}{9}-\frac{32}{9}$

= $\frac{76}{9}$

= ${\left[-\frac{1}{2}{x}^{-2}+\frac{9}{4}\cdot \cos \left(x\right)\right]}_{\pi }^{\frac{3}{2}\pi }$

= ${\left[-\frac{1}{2x^2}+\frac{9}{4}\cdot \cos \left(x\right)\right]}_{\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{1}{2{\left(\frac{3}{2}\pi \right)}^2}+\frac{9}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(-\frac{1}{2{\pi }^2}+\frac{9}{4}\cdot \cos \left(\pi \right)\right)$

= $-\frac{1}{2{\left(\frac{3}{2}\pi \right)}^2}+\frac{9}{4}\cdot 0-\left(-\frac{1}{2{\pi }^2}+\frac{9}{4}\cdot \left(-1\right)\right)$

= $-\frac{1}{2{\left(\frac{3}{2}\pi \right)}^2}+0-\left(-\frac{1}{2{\pi }^2}-\frac{9}{4}\right)$

= $-\frac{2}{9{\pi }^2}-\left(-\frac{9}{4}-\frac{1}{2{\pi }^2}\right)$

= $-1·\left(-\frac{9}{4}\right)-1·\left(-\frac{1}{2{\pi }^2}\right)-\frac{2}{9{\pi }^2}$

= $\frac{9}{4}+\frac{1}{2{\pi }^2}-\frac{2}{9{\pi }^2}$

= $\frac{9}{4}+\frac{5}{18{\pi }^2}$

= ${\left[-3e^{x-3}\right]}_0^3$

$=$ $-3e^{3-3}+3e^{0-3}$

= $-3e^0+3e^{-3}$

= $-3+3e^{-3}$