$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 6 -1 -3 = 2

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

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

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

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

= $\frac{1}{3}+\frac{9}{3}-\left(\frac{8}{3}+\frac{18}{3}\right)$

= $\frac{10}{3}-1·\frac{26}{3}$

= $\frac{10}{3}-\frac{26}{3}$

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

= ${\left[-\frac{1}{2}{x}^{\frac{5}{2}}-\frac{1}{2}{x}^{-2}\right]}_4^9$

= ${\left[-\frac{1}{2}{\left(\sqrt{x}\right)}^5-\frac{1}{2x^2}\right]}_4^9$

$=$ $-\frac{1}{2}{\left(\sqrt{9}\right)}^5-\frac{1}{2\cdot 9^2}-\left(-\frac{1}{2}{\left(\sqrt{4}\right)}^5-\frac{1}{2\cdot 4^2}\right)$

= $-\frac{1}{2}\cdot 3^5-\frac{1}{2}\cdot \left(\frac{1}{81}\right)-\left(-\frac{1}{2}\cdot 2^5-\frac{1}{2}\cdot \left(\frac{1}{16}\right)\right)$

= $-\frac{1}{2}\cdot 243-\frac{1}{162}-\left(-\frac{1}{2}\cdot 32-\frac{1}{32}\right)$

= $-\frac{243}{2}-\frac{1}{162}-\left(-16-\frac{1}{32}\right)$

= $-\frac{19683}{162}-\frac{1}{162}-\left(-\frac{512}{32}-\frac{1}{32}\right)$

= $-\frac{9842}{81}-1·\left(-\frac{513}{32}\right)$

= $-\frac{9842}{81}+\frac{513}{32}$

= $-\frac{314944}{2592}+\frac{41553}{2592}$

= $-\frac{9842}{81}+\frac{513}{32}$

= $-\frac{314944}{2592}+\frac{41553}{2592}$

= $-\frac{273391}{2592}$

= ${\left[\frac{1}{6}{\left(-3x+5\right)}^{-2}\right]}_3^6$

= ${\left[\frac{1}{6{\left(-3x+5\right)}^2}\right]}_3^6$

$=$ $\frac{1}{6{\left(-3\cdot 6+5\right)}^2}-\frac{1}{6{\left(-3\cdot 3+5\right)}^2}$

= $\frac{1}{6{\left(-18+5\right)}^2}-\frac{1}{6{\left(-9+5\right)}^2}$

= $\frac{1}{6{\left(-13\right)}^2}-\frac{1}{6{\left(-4\right)}^2}$

= $\frac{1}{6}\cdot \left(\frac{1}{169}\right)-\frac{1}{6}\cdot \left(\frac{1}{16}\right)$

= $\frac{1}{1014}-\frac{1}{96}$

= $\frac{16}{16224}-\frac{169}{16224}$

= $-\frac{51}{5408}$

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

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

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

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

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

= $-5-\frac{4}{3\pi }+\frac{2}{\pi }$

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

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

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

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

= $3\cdot 0-3\cdot 1$

= $0-3$

= $-3$