$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 1 = 3 ⋅ 1 = 3.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 1 +3 = 4

= ${\left[-\frac{1}{3}{x}^3+\frac{1}{2}{x}^2\right]}_{-1}^0$

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

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

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

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

= $-1·\frac{5}{6}$

= $-\frac{5}{6}$

= ${\left[\frac{7}{6}{x}^3+6\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $\frac{7}{6}\cdot {\left(\frac{1}{2}\pi \right)}^3+6\cdot \sin \left(\frac{1}{2}\pi \right)-\left(\frac{7}{6}\cdot {\left(0\right)}^3+6\cdot \sin \left(0\right)\right)$

= $\frac{7}{6}\cdot {\left(\frac{1}{2}\pi \right)}^3+6\cdot 1-\left(\frac{7}{6}\cdot 0+6\cdot 0\right)$

= $\frac{7}{6}\cdot {\left(\frac{1}{2}\pi \right)}^3+6-\left(0+0\right)$

= $6+\frac{7}{48}\cdot {\pi }^3+0$

= $6+\frac{7}{48}\cdot {\pi }^3$

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

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

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

= $-\frac{2}{3}\cdot 1+\frac{2}{3}\cdot 0$

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

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

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

= ${\left[-\frac{4}{3}{e}^{-3x}+\frac{2}{3}{x}^{\frac{3}{2}}\right]}_4^9$

= ${\left[-\frac{4}{3}{e}^{-3x}+\frac{2}{3}{\left(\sqrt{x}\right)}^3\right]}_4^9$

$=$ $-\frac{4}{3}{e}^{-3\cdot 9}+\frac{2}{3}{\left(\sqrt{9}\right)}^3-\left(-\frac{4}{3}{e}^{-3\cdot 4}+\frac{2}{3}{\left(\sqrt{4}\right)}^3\right)$

= $-\frac{4}{3}{e}^{-27}+\frac{2}{3}\cdot 3^3-\left(-\frac{4}{3}{e}^{-12}+\frac{2}{3}\cdot 2^3\right)$

= $-\frac{4}{3}{e}^{-27}+\frac{2}{3}\cdot 27-\left(-\frac{4}{3}{e}^{-12}+\frac{2}{3}\cdot 8\right)$

= $-\frac{4}{3}{e}^{-27}+18-\left(-\frac{4}{3}{e}^{-12}+\frac{16}{3}\right)$

= $\frac{4}{3}{e}^{-12}-1·\frac{16}{3}-\frac{4}{3}{e}^{-27}+18$

= $\frac{4}{3}{e}^{-12}-\frac{16}{3}-\frac{4}{3}{e}^{-27}+18$

= $\frac{4}{3}{e}^{-12}-\frac{4}{3}{e}^{-27}-\frac{16}{3}+18$

= $\frac{4}{3}{e}^{-12}-\frac{4}{3}{e}^{-27}+\frac{38}{3}$

= ${\left[\frac{2}{9}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

= ${\left[\frac{2}{9}{\left(\sqrt{3x-6}\right)}^3\right]}_{\frac{22}{3}}^{\frac{31}{3}}$

$=$ $\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{31}{3}\right)-6}\right)}^3-\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{31-6}\right)}^3-\frac{2}{9}{\left(\sqrt{22-6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{25}\right)}^3-\frac{2}{9}{\left(\sqrt{16}\right)}^3$

= $\frac{2}{9}\cdot 5^3-\frac{2}{9}\cdot 4^3$

= $\frac{2}{9}\cdot 125-\frac{2}{9}\cdot 64$

= $\frac{250}{9}-\frac{128}{9}$

= $\frac{122}{9}$