$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 1 = 3 ⋅ 1 = 3.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4.5 +1 +3 = -0.5

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

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

= $-\frac{2}{3}\cdot 8-6-\left(-\frac{2}{3}\cdot 0+0\right)$

= $-\frac{16}{3}-6-\left(0+0\right)$

= $-\frac{16}{3}-\frac{18}{3}+0$

= $-\frac{34}{3}$

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

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

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

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

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

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

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

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

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

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

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

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

$=$ $-\frac{2}{3}{\left(\sqrt{3\cdot 4-3}\right)}^3+\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{12-3}\right)}^3+\frac{2}{3}{\left(\sqrt{7-3}\right)}^3$

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

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

= $-\frac{2}{3}\cdot 27+\frac{2}{3}\cdot 8$

= $-18+\frac{16}{3}$

= $-\frac{54}{3}+\frac{16}{3}$

= $-\frac{38}{3}$

= ${\left[\frac{1}{6}{x}^6-\frac{8}{3}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{1}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{8}{3}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(\frac{1}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{8}{3}\cdot \left(-1\right)-\left(\frac{1}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\cdot 1\right)$

= $\frac{1}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{8}{3}-\left(\frac{1}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\right)$

= $\frac{8}{3}+\frac{243}{128}\cdot {\pi }^6-\left(-\frac{8}{3}+\frac{1}{384}\cdot {\pi }^6\right)$

= $\frac{8}{3}+\frac{243}{128}\cdot {\pi }^6-1·\left(-\frac{8}{3}\right)-1·\frac{1}{384}\cdot {\pi }^6$

= $\frac{8}{3}+\frac{243}{128}\cdot {\pi }^6+\frac{8}{3}-\frac{1}{384}\cdot {\pi }^6$

= $\frac{8}{3}+\frac{8}{3}+\frac{243}{128}\cdot {\pi }^6-\frac{1}{384}\cdot {\pi }^6$

= $\frac{16}{3}+\frac{91}{48}\cdot {\pi }^6$

= ${\left[\frac{4}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[\frac{4}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $\frac{4}{3}{\left(\sqrt{27-2}\right)}^3-\frac{4}{3}{\left(\sqrt{18-2}\right)}^3$

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

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

= $\frac{4}{3}\cdot 125-\frac{4}{3}\cdot 64$

= $\frac{500}{3}-\frac{256}{3}$

= $\frac{244}{3}$