$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -9 -4.5 +4.5 = -9

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

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

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

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

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

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

= ${\left[-6\cdot \cos \left(x\right)-\frac{1}{2}{x}^4\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $-6\cdot \left(-1\right)-\frac{1}{2}\cdot {\pi }^4-\left(-6\cdot 0-\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $6-\frac{1}{2}\cdot {\pi }^4-\left(0-\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $6-\frac{1}{2}\cdot {\pi }^4+\frac{1}{32}\cdot {\pi }^4$

= $6-\frac{15}{32}\cdot {\pi }^4$

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

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

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

= $-0-1$

= $-1$

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

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

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

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

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

= $\frac{5}{2\pi }-\left(-\frac{3}{2}+\frac{10}{\pi }\right)$

= $-1·\left(-\frac{3}{2}\right)-1·\frac{10}{\pi }+\frac{5}{2\pi }$

= $\frac{3}{2}-\frac{10}{\pi }+\frac{5}{2\pi }$

= $\frac{3}{2}-\frac{15}{2\pi }$

= ${\left[-\frac{2}{3}{\left(-x+2\right)}^3+\frac{1}{2}{x}^2\right]}_2^3$

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

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

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

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

= $\frac{4}{6}+\frac{27}{6}-2$

= $\frac{31}{6}-2$

= $\frac{31}{6}-\frac{12}{6}$

= $\frac{19}{6}$