$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

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 = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -12 -6 +4.5 = -13.5

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

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

= $-1+1-\left(-\left(-8\right)+4\right)$

= $-1+1-\left(8+4\right)$

= $0-1·12$

= $0-12$

= $-12$

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

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

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

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

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

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

= $1+\frac{5}{6\pi }-1·\left(-1\right)-1·\frac{5}{2\pi }$

= $1+\frac{5}{6\pi }+1-\frac{5}{2\pi }$

= $1+1+\frac{5}{6\pi }-\frac{5}{2\pi }$

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

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

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

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

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

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

= $-\frac{1}{3}+\frac{27}{3}-\left(\frac{1}{3}+\frac{9}{3}\right)$

= $\frac{26}{3}-1·\frac{10}{3}$

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

= $\frac{16}{3}$

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

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

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

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

= $-\frac{1}{3}{e}^{-\frac{9}{2}\pi }-\frac{1}{2}+\frac{1}{3}{e}^{-\frac{3}{2}\pi }-1·\frac{1}{2}$

= $-\frac{1}{3}{e}^{-\frac{9}{2}\pi }-\frac{1}{2}+\frac{1}{3}{e}^{-\frac{3}{2}\pi }-\frac{1}{2}$

= $\frac{1}{3}{e}^{-\frac{3}{2}\pi }-\frac{1}{3}{e}^{-\frac{9}{2}\pi }-\frac{1}{2}-\frac{1}{2}$

= $\frac{1}{3}{e}^{-\frac{3}{2}\pi }-\frac{1}{3}{e}^{-\frac{9}{2}\pi }-1$

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

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

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

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

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

= $-1$