$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

Somit gilt:

$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -3 +1 = -2

= ${\left[-\frac{4}{3}{x}^3+5x\right]}_{-3}^0$

$=$ $-\frac{4}{3}\cdot 0^3+5\cdot 0-\left(-\frac{4}{3}\cdot {\left(-3\right)}^3+5\cdot \left(-3\right)\right)$

= $-\frac{4}{3}\cdot 0+0-\left(-\frac{4}{3}\cdot \left(-27\right)-15\right)$

= $0+0-\left(36-15\right)$

= $0-1·21$

= $-21$

= ${\left[-\frac{5}{6}{e}^{-3x}-\frac{3}{8}{x}^6\right]}_0^3$

$=$ $-\frac{5}{6}{e}^{-3\cdot 3}-\frac{3}{8}\cdot 3^6-\left(-\frac{5}{6}{e}^{-3\cdot 0}-\frac{3}{8}\cdot 0^6\right)$

= $-\frac{5}{6}{e}^{-9}-\frac{3}{8}\cdot 729-\left(-\frac{5}{6}{e}^0-\frac{3}{8}\cdot 0\right)$

= $-\frac{5}{6}{e}^{-9}-\frac{2187}{8}-\left(-\frac{5}{6}+0\right)$

= $-\frac{5}{6}{e}^{-9}-\frac{2187}{8}-\left(-\frac{5}{6}+0\right)$

= $-\frac{5}{6}{e}^{-9}-\frac{2187}{8}+\frac{5}{6}$

= $-\frac{5}{6}{e}^{-9}-\frac{6541}{24}$

= ${\left[-\frac{4}{3}{\left(-x+2\right)}^{\frac{3}{2}}\right]}_{-14}^{-7}$

= ${\left[-\frac{4}{3}{\left(\sqrt{-x+2}\right)}^3\right]}_{-14}^{-7}$

$=$ $-\frac{4}{3}{\left(\sqrt{-\left(-7\right)+2}\right)}^3+\frac{4}{3}{\left(\sqrt{-\left(-14\right)+2}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{7+2}\right)}^3+\frac{4}{3}{\left(\sqrt{14+2}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{9}\right)}^3+\frac{4}{3}{\left(\sqrt{16}\right)}^3$

= $-\frac{4}{3}\cdot 3^3+\frac{4}{3}\cdot 4^3$

= $-\frac{4}{3}\cdot 27+\frac{4}{3}\cdot 64$

= $-36+\frac{256}{3}$

= $-\frac{108}{3}+\frac{256}{3}$

= $\frac{148}{3}$

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

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

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

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

= $0-\frac{3}{2}+3$

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

= $-\frac{3}{2}+\frac{6}{2}$

= $\frac{3}{2}$

= ${\left[{\left(2x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{2}}^{\frac{19}{2}}$

= ${\left[{\left(\sqrt{2x-3}\right)}^3\right]}_{\frac{7}{2}}^{\frac{19}{2}}$

$=$ ${\left(\sqrt{2\cdot \left(\frac{19}{2}\right)-3}\right)}^3-{\left(\sqrt{2\cdot \left(\frac{7}{2}\right)-3}\right)}^3$

= ${\left(\sqrt{19-3}\right)}^3-{\left(\sqrt{7-3}\right)}^3$

= ${\left(\sqrt{16}\right)}^3-{\left(\sqrt{4}\right)}^3$

= $4^3-2^3$

= $64-8$

= $56$