$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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.

Somit gilt:

$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -12 -6 = -18

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

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

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

= $1+\frac{3}{2}-\left(-8+6\right)$

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

= $\frac{5}{2}+2$

= $\frac{5}{2}+\frac{4}{2}$

= $\frac{9}{2}$

= ${\left[-\frac{1}{4}\cdot \sin \left(x\right)-\frac{1}{2}{e}^x\right]}_0^{\frac{3}{2}\pi }$

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

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

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

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

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

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

= ${\left[e^{-3x+5}\right]}_1^4$

$=$ $e^{-3\cdot 4+5}-e^{-3\cdot 1+5}$

= $e^{-12+5}-e^{-3+5}$

= $e^{-7}-e^2$

= ${\left[-\frac{5}{18}{x}^6-\frac{8}{3}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{5}{18}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{8}{3}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{5}{18}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{18}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{8}{3}\cdot \left(-1\right)-\left(-\frac{5}{18}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\cdot 1\right)$

= $-\frac{5}{18}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{8}{3}-\left(-\frac{5}{18}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{8}{3}\right)$

= $\frac{8}{3}-\frac{405}{128}\cdot {\pi }^6-\left(-\frac{8}{3}-\frac{5}{1152}\cdot {\pi }^6\right)$

= $\frac{8}{3}-\frac{405}{128}\cdot {\pi }^6-1·\left(-\frac{8}{3}\right)-1·\left(-\frac{5}{1152}\cdot {\pi }^6\right)$

= $\frac{8}{3}-\frac{405}{128}\cdot {\pi }^6+\frac{8}{3}+\frac{5}{1152}\cdot {\pi }^6$

= $\frac{8}{3}+\frac{8}{3}-\frac{405}{128}\cdot {\pi }^6+\frac{5}{1152}\cdot {\pi }^6$

= $\frac{16}{3}-\frac{455}{144}\cdot {\pi }^6$

= ${\left[3e^{-x+3}\right]}_0^3$

$=$ $3e^{-3+3}-3e^{-0+3}$

= $3e^0-3e^3$

= $3-3e^3$