$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 6) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 1.5 +2 = 3.5

= ${\left[-\frac{2}{3}{x}^3+\frac{5}{2}{x}^2\right]}_{-2}^0$

$=$ $-\frac{2}{3}\cdot 0^3+\frac{5}{2}\cdot 0^2-\left(-\frac{2}{3}\cdot {\left(-2\right)}^3+\frac{5}{2}\cdot {\left(-2\right)}^2\right)$

= $-\frac{2}{3}\cdot 0+\frac{5}{2}\cdot 0-\left(-\frac{2}{3}\cdot \left(-8\right)+\frac{5}{2}\cdot 4\right)$

= $0+0-\left(\frac{16}{3}+10\right)$

= $0-\left(\frac{16}{3}+\frac{30}{3}\right)$

= $-1·\frac{46}{3}$

= $-\frac{46}{3}$

= ${\left[-\frac{1}{3}{x}^3-\frac{5}{3}{e}^{3x}\right]}_1^5$

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

= $-\frac{1}{3}\cdot 125-\frac{5}{3}{e}^{15}-\left(-\frac{1}{3}\cdot 1-\frac{5}{3}{e}^3\right)$

= $-\frac{125}{3}-\frac{5}{3}{e}^{15}-\left(-\frac{1}{3}-\frac{5}{3}{e}^3\right)$

= $-\frac{5}{3}{e}^{15}-\frac{125}{3}-\left(-\frac{5}{3}{e}^3-\frac{1}{3}\right)$

= $-\frac{5}{3}{e}^{15}-\frac{125}{3}+\frac{5}{3}{e}^3-1·\left(-\frac{1}{3}\right)$

= $-\frac{5}{3}{e}^{15}-\frac{125}{3}+\frac{5}{3}{e}^3+\frac{1}{3}$

= $-\frac{5}{3}{e}^{15}+\frac{5}{3}{e}^3-\frac{125}{3}+\frac{1}{3}$

= $-\frac{5}{3}{e}^{15}+\frac{5}{3}{e}^3-\frac{124}{3}$

= ${\left[2{\left(-x+1\right)}^{\frac{3}{2}}\right]}_{-15}^{-8}$

= ${\left[2{\left(\sqrt{-x+1}\right)}^3\right]}_{-15}^{-8}$

$=$ $2{\left(\sqrt{-\left(-8\right)+1}\right)}^3-2{\left(\sqrt{-\left(-15\right)+1}\right)}^3$

= $2{\left(\sqrt{8+1}\right)}^3-2{\left(\sqrt{15+1}\right)}^3$

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

= $2\cdot 3^3-2\cdot 4^3$

= $2\cdot 27-2\cdot 64$

= $54-128$

= $-74$

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

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

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

= $2\cdot 0+\frac{5}{3{\left(2\pi \right)}^3}-\left(2\cdot 1+\frac{5}{3{\left(\frac{1}{2}\pi \right)}^3}\right)$

= $0+\frac{5}{3{\left(2\pi \right)}^3}-\left(2+\frac{5}{3{\left(\frac{1}{2}\pi \right)}^3}\right)$

= $\frac{5}{24{\pi }^3}-\left(2+\frac{40}{3{\pi }^3}\right)$

= $-1·2-1·\frac{40}{3{\pi }^3}+\frac{5}{24{\pi }^3}$

= $-2-\frac{40}{3{\pi }^3}+\frac{5}{24{\pi }^3}$

= $-2-\frac{105}{8{\pi }^3}$

= ${\left[\frac{2}{3}{e}^{-3x+5}\right]}_2^3$

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

= $\frac{2}{3}{e}^{-9+5}-\frac{2}{3}{e}^{-6+5}$

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