$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -6 +1.5 = -4.5

= ${\left[-\frac{5}{2}{x}^2+4x\right]}_{-2}^1$

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

= $-\frac{5}{2}\cdot 1+4-\left(-\frac{5}{2}\cdot 4-8\right)$

= $-\frac{5}{2}+4-\left(-10-8\right)$

= $-\frac{5}{2}+\frac{8}{2}-1·\left(-18\right)$

= $\frac{3}{2}+18$

= $\frac{3}{2}+\frac{36}{2}$

= $\frac{39}{2}$

= ${\left[-\frac{8}{3}{x}^3-\frac{4}{3}{x}^{\frac{3}{2}}\right]}_4^9$

= ${\left[-\frac{8}{3}{x}^3-\frac{4}{3}{\left(\sqrt{x}\right)}^3\right]}_4^9$

$=$ $-\frac{8}{3}\cdot 9^3-\frac{4}{3}{\left(\sqrt{9}\right)}^3-\left(-\frac{8}{3}\cdot 4^3-\frac{4}{3}{\left(\sqrt{4}\right)}^3\right)$

= $-\frac{8}{3}\cdot 729-\frac{4}{3}\cdot 3^3-\left(-\frac{8}{3}\cdot 64-\frac{4}{3}\cdot 2^3\right)$

= $-1944-\frac{4}{3}\cdot 27-\left(-\frac{512}{3}-\frac{4}{3}\cdot 8\right)$

= $-1944-36-\left(-\frac{512}{3}-\frac{32}{3}\right)$

= $-1980-1·\left(-\frac{544}{3}\right)$

= $-1980+\frac{544}{3}$

= $-\frac{5940}{3}+\frac{544}{3}$

= $-\frac{5396}{3}$

= ${\left[-\frac{2}{3}{\left(-2x+4\right)}^{\frac{3}{2}}\right]}_{-\frac{21}{2}}^{-6}$

= ${\left[-\frac{2}{3}{\left(\sqrt{-2x+4}\right)}^3\right]}_{-\frac{21}{2}}^{-6}$

$=$ $-\frac{2}{3}{\left(\sqrt{-2\cdot \left(-6\right)+4}\right)}^3+\frac{2}{3}{\left(\sqrt{-2\cdot \left(-\frac{21}{2}\right)+4}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{12+4}\right)}^3+\frac{2}{3}{\left(\sqrt{21+4}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{16}\right)}^3+\frac{2}{3}{\left(\sqrt{25}\right)}^3$

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

= $-\frac{2}{3}\cdot 64+\frac{2}{3}\cdot 125$

= $-\frac{128}{3}+\frac{250}{3}$

= $\frac{122}{3}$

= ${\left[\frac{1}{5}{x}^5+\frac{4}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{1}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5+\frac{4}{3}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\frac{1}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5+\frac{4}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5+\frac{4}{3}\cdot 0-\left(\frac{1}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5+\frac{4}{3}\cdot 0\right)$

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

= $\frac{243}{160}\cdot {\pi }^5-\frac{1}{160}\cdot {\pi }^5$

= $\frac{121}{80}\cdot {\pi }^5$

= ${\left[-\frac{1}{2}{\left(-2x+2\right)}^3\right]}_1^3$

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

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

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

= $-\frac{1}{2}\cdot \left(-64\right)+\frac{1}{2}\cdot 0$

= $32+0$

= $32$