$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 4 = 2 ⋅ 4 = 8.

Somit gilt:

$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -3 +4 +8 = 9

= ${\left[\frac{2}{3}{x}^3-5x\right]}_{-3}^{-2}$

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

= $\frac{2}{3}\cdot \left(-8\right)+10-\left(\frac{2}{3}\cdot \left(-27\right)+15\right)$

= $-\frac{16}{3}+10-\left(-18+15\right)$

= $-\frac{16}{3}+\frac{30}{3}-1·\left(-3\right)$

= $\frac{14}{3}+3$

= $\frac{14}{3}+\frac{9}{3}$

= $\frac{23}{3}$

= ${\left[-\frac{1}{6}{e}^{-2x}+\frac{1}{6}{x}^{\frac{3}{2}}\right]}_1^4$

= ${\left[-\frac{1}{6}{e}^{-2x}+\frac{1}{6}{\left(\sqrt{x}\right)}^3\right]}_1^4$

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

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

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

= $-\frac{1}{6}{e}^{-8}+\frac{4}{3}-\left(-\frac{1}{6}{e}^{-2}+\frac{1}{6}\right)$

= $\frac{1}{6}{e}^{-2}-1·\frac{1}{6}-\frac{1}{6}{e}^{-8}+\frac{4}{3}$

= $\frac{1}{6}{e}^{-2}-\frac{1}{6}-\frac{1}{6}{e}^{-8}+\frac{4}{3}$

= $\frac{1}{6}{e}^{-2}-\frac{1}{6}{e}^{-8}-\frac{1}{6}+\frac{4}{3}$

= $\frac{1}{6}{e}^{-2}-\frac{1}{6}{e}^{-8}+\frac{7}{6}$

= ${\left[\frac{2}{3}{\left(x-1\right)}^{-3}\right]}_2^5$

= ${\left[\frac{2}{3{\left(x-1\right)}^3}\right]}_2^5$

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

= $\frac{2}{3\cdot 4^3}-\frac{2}{3\cdot 1^3}$

= $\frac{2}{3}\cdot \left(\frac{1}{64}\right)-\frac{2}{3}\cdot 1$

= $\frac{1}{96}-\frac{2}{3}$

= $\frac{1}{96}-\frac{64}{96}$

= $-\frac{21}{32}$

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

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

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

= $-\frac{9}{16}\cdot {\pi }^3+\frac{1}{48}\cdot {\pi }^3$

= $-\frac{13}{24}\cdot {\pi }^3$

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

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

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

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

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

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