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

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

Somit gilt:

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

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

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

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

= $\frac{5}{2}-1-\left(10-2\right)$

= $\frac{5}{2}-\frac{2}{2}-1·8$

= $\frac{3}{2}-8$

= $\frac{3}{2}-\frac{16}{2}$

= $-\frac{13}{2}$

= ${\left[\frac{3}{4}{x}^6+\frac{5}{9}{e}^{3x}\right]}_{-2}^1$

$=$ $\frac{3}{4}\cdot 1^6+\frac{5}{9}{e}^{3\cdot 1}-\left(\frac{3}{4}\cdot {\left(-2\right)}^6+\frac{5}{9}{e}^{3\cdot \left(-2\right)}\right)$

= $\frac{3}{4}\cdot 1+\frac{5}{9}{e}^3-\left(\frac{3}{4}\cdot 64+\frac{5}{9}{e}^{-6}\right)$

= $\frac{3}{4}+\frac{5}{9}{e}^3-\left(48+\frac{5}{9}{e}^{-6}\right)$

= $\frac{5}{9}{e}^3+\frac{3}{4}-\left(\frac{5}{9}{e}^{-6}+48\right)$

= $\frac{5}{9}{e}^3+\frac{3}{4}-\frac{5}{9}{e}^{-6}-1·48$

= $\frac{5}{9}{e}^3+\frac{3}{4}-\frac{5}{9}{e}^{-6}-48$

= $\frac{5}{9}{e}^3-\frac{5}{9}{e}^{-6}+\frac{3}{4}-48$

= $\frac{5}{9}{e}^3-\frac{5}{9}{e}^{-6}-\frac{189}{4}$

= ${\left[-\frac{1}{9}{\left(-3x+6\right)}^3-\frac{3}{2}{x}^2\right]}_1^2$

$=$ $-\frac{1}{9}\cdot {\left(-3\cdot 2+6\right)}^3-\frac{3}{2}\cdot 2^2-\left(-\frac{1}{9}\cdot {\left(-3\cdot 1+6\right)}^3-\frac{3}{2}\cdot 1^2\right)$

= $-\frac{1}{9}\cdot {\left(-6+6\right)}^3-\frac{3}{2}\cdot 4-\left(-\frac{1}{9}\cdot {\left(-3+6\right)}^3-\frac{3}{2}\cdot 1\right)$

= $-\frac{1}{9}\cdot 0^3-6-\left(-\frac{1}{9}\cdot 3^3-\frac{3}{2}\right)$

= $-\frac{1}{9}\cdot 0-6-\left(-\frac{1}{9}\cdot 27-\frac{3}{2}\right)$

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

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

= $-6-1·\left(-\frac{9}{2}\right)$

= $-6+\frac{9}{2}$

= $-\frac{12}{2}+\frac{9}{2}$

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

= ${\left[2\cdot \cos \left(x\right)+\sin \left(x\right)\right]}_0^{\pi }$

$=$ $2\cdot \cos \left(\pi \right)+\sin \left(\pi \right)-\left(2\cdot \cos \left(0\right)+\sin \left(0\right)\right)$

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

= $-2+0-\left(2+0\right)$

= $-2-2$

= $-4$

= ${\left[\frac{2}{9}{\left(3x-6\right)}^3+6x\right]}_2^3$

$=$ $\frac{2}{9}\cdot {\left(3\cdot 3-6\right)}^3+6\cdot 3-\left(\frac{2}{9}\cdot {\left(3\cdot 2-6\right)}^3+6\cdot 2\right)$

= $\frac{2}{9}\cdot {\left(9-6\right)}^3+18-\left(\frac{2}{9}\cdot {\left(6-6\right)}^3+12\right)$

= $\frac{2}{9}\cdot 3^3+18-\left(\frac{2}{9}\cdot 0^3+12\right)$

= $\frac{2}{9}\cdot 27+18-\left(\frac{2}{9}\cdot 0+12\right)$

= $6+18-\left(0+12\right)$

= $24-12$

= $12$