$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 2 = 3 ⋅ 2 = 6.

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

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

Somit gilt:

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

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

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

= $-\frac{2}{3}\cdot \left(-8\right)-\frac{1}{2}\cdot 4-\left(-\frac{2}{3}\cdot \left(-27\right)-\frac{1}{2}\cdot 9\right)$

= $\frac{16}{3}-2-\left(18-\frac{9}{2}\right)$

= $\frac{16}{3}-\frac{6}{3}-\left(\frac{36}{2}-\frac{9}{2}\right)$

= $\frac{10}{3}-1·\frac{27}{2}$

= $\frac{10}{3}-\frac{27}{2}$

= $\frac{20}{6}-\frac{81}{6}$

= $-\frac{61}{6}$

= ${\left[-\frac{1}{4}{e}^{-x}+\frac{3}{4}{x}^{-2}\right]}_2^4$

= ${\left[-\frac{1}{4}{e}^{-x}+\frac{3}{4x^2}\right]}_2^4$

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

= $-\frac{1}{4}{e}^{-4}+\frac{3}{4}\cdot \left(\frac{1}{16}\right)-\left(-\frac{1}{4}{e}^{-2}+\frac{3}{4}\cdot \left(\frac{1}{4}\right)\right)$

= $-\frac{1}{4}{e}^{-4}+\frac{3}{64}-\left(-\frac{1}{4}{e}^{-2}+\frac{3}{16}\right)$

= $\frac{1}{4}{e}^{-2}-1·\frac{3}{16}-\frac{1}{4}{e}^{-4}+\frac{3}{64}$

= $\frac{1}{4}{e}^{-2}-\frac{3}{16}-\frac{1}{4}{e}^{-4}+\frac{3}{64}$

= $\frac{1}{4}{e}^{-2}-\frac{1}{4}{e}^{-4}-\frac{3}{16}+\frac{3}{64}$

= $\frac{1}{4}{e}^{-2}-\frac{1}{4}{e}^{-4}-\frac{9}{64}$

= ${\left[-e^{-x+1}\right]}_2^3$

$=$ $-e^{-3+1}+e^{-2+1}$

= $-e^{-2}+e^{-1}$

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

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

= $2\cdot 0-4\cdot \left(-1\right)-\left(2\cdot 1-4\cdot 0\right)$

= $0+4-\left(2+0\right)$

= $4-2$

= $2$

= ${\left[\frac{1}{2}\cdot \sin \left(-2x-\frac{3}{2}\pi \right)\right]}_0^{\frac{3}{2}\pi }$

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

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

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

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

= $-1$