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

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

Somit gilt:

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

= ${\left[-\frac{4}{3}{x}^3-4x\right]}_{-3}^1$

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

= $-\frac{4}{3}\cdot 1-4-\left(-\frac{4}{3}\cdot \left(-27\right)+12\right)$

= $-\frac{4}{3}-4-\left(36+12\right)$

= $-\frac{4}{3}-\frac{12}{3}-1·48$

= $-\frac{16}{3}-48$

= $-\frac{16}{3}-\frac{144}{3}$

= $-\frac{160}{3}$

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

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

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

= $-\frac{1}{3}\cdot {\pi }^6-1·\left(-1\right)-1·\left(-\frac{1}{192}\cdot {\pi }^6\right)$

= $-\frac{1}{3}\cdot {\pi }^6+1+\frac{1}{192}\cdot {\pi }^6$

= $1-\frac{1}{3}\cdot {\pi }^6+\frac{1}{192}\cdot {\pi }^6$

= $1-\frac{21}{64}\cdot {\pi }^6$

= ${\left[-{\left(-3x+5\right)}^{-1}\right]}_2^3$

= ${\left[-\frac{1}{-3x+5}\right]}_2^3$

$=$ $-\frac{1}{-3\cdot 3+5}+\frac{1}{-3\cdot 2+5}$

= $-\frac{1}{-9+5}+\frac{1}{-6+5}$

= $-\frac{1}{\left(-4\right)}+\frac{1}{\left(-1\right)}$

= $-\left(-\frac{1}{4}\right)+\left(-1\right)$

= $\frac{1}{4}-1$

= $\frac{1}{4}-\frac{4}{4}$

= $-\frac{3}{4}$

= ${\left[-7\cdot \sin \left(x\right)-\frac{7}{15}{x}^5\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-7\cdot \sin \left(\frac{3}{2}\pi \right)-\frac{7}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(-7\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{7}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $-7\cdot \left(-1\right)-\frac{7}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(-7\cdot 1-\frac{7}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $7-\frac{7}{15}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(-7-\frac{7}{15}\cdot {\left(\frac{1}{2}\pi \right)}^5\right)$

= $7-\frac{567}{160}\cdot {\pi }^5-\left(-7-\frac{7}{480}\cdot {\pi }^5\right)$

= $7-\frac{567}{160}\cdot {\pi }^5-1·\left(-7\right)-1·\left(-\frac{7}{480}\cdot {\pi }^5\right)$

= $7-\frac{567}{160}\cdot {\pi }^5+7+\frac{7}{480}\cdot {\pi }^5$

= $7+7-\frac{567}{160}\cdot {\pi }^5+\frac{7}{480}\cdot {\pi }^5$

= $14-\frac{847}{240}\cdot {\pi }^5$

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

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

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

= $-\frac{1}{3}\cdot \left(-1\right)-\frac{1}{2}-\left(-\frac{1}{3}\cdot \left(-8\right)+0\right)$

= $\frac{1}{3}-\frac{1}{2}-\left(\frac{8}{3}+0\right)$

= $\frac{2}{6}-\frac{3}{6}-\left(\frac{8}{3}+0\right)$

= $-\frac{1}{6}-\frac{8}{3}$

= $-\frac{1}{6}-\frac{16}{6}$

= $-\frac{17}{6}$