$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : 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.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4 -8 -5 = -17

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

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

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

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

= $\frac{2}{6}-\frac{3}{6}-\left(\frac{8}{3}-\frac{6}{3}\right)$

= $-\frac{1}{6}-1·\frac{2}{3}$

= $-\frac{1}{6}-\frac{2}{3}$

= $-\frac{1}{6}-\frac{4}{6}$

= $-\frac{5}{6}$

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

$=$ $-\frac{5}{3}{e}^{3\cdot \pi }-\sin \left(\pi \right)-\left(-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\sin \left(\frac{1}{2}\pi \right)\right)$

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

= $-\frac{5}{3}{e}^{3\pi }+\frac{5}{3}{e}^{\frac{3}{2}\pi }-1·\left(-1\right)$

= $-\frac{5}{3}{e}^{3\pi }+\frac{5}{3}{e}^{\frac{3}{2}\pi }+1$

= ${\left[\frac{4}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_5^{10}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-1}\right)}^3\right]}_5^{10}$

$=$ $\frac{4}{3}{\left(\sqrt{10-1}\right)}^3-\frac{4}{3}{\left(\sqrt{5-1}\right)}^3$

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

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

= $\frac{4}{3}\cdot 27-\frac{4}{3}\cdot 8$

= $36-\frac{32}{3}$

= $\frac{108}{3}-\frac{32}{3}$

= $\frac{76}{3}$

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

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

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

= $-\frac{1}{2}+0-\left(0-2\right)$

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

= $-\frac{1}{2}+2$

= $-\frac{1}{2}+\frac{4}{2}$

= $\frac{3}{2}$

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

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

= $-\frac{1}{3}\cdot {\left(-6+3\right)}^3+\frac{1}{3}\cdot {\left(0+3\right)}^3$

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

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

= $9+9$

= $18$