$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 3 = 3 ⋅ 3 = 9.

Somit gilt:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 4.5 +9 = 13.5

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

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

= $-\frac{5}{2}\cdot 4+10-\left(-\frac{5}{2}\cdot 9+15\right)$

= $-10+10-\left(-\frac{45}{2}+15\right)$

= $0-\left(-\frac{45}{2}+\frac{30}{2}\right)$

= $0-1·\left(-\frac{15}{2}\right)$

= $0+\frac{15}{2}$

= $0+\frac{15}{2}$

= $\frac{15}{2}$

= ${\left[\frac{1}{3}{x}^6+5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{1}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6+5\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\frac{1}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6+5\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6+5\cdot 0-\left(\frac{1}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6+5\cdot 0\right)$

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

= $\frac{243}{64}\cdot {\pi }^6-\frac{1}{192}\cdot {\pi }^6$

= $\frac{729}{192}\cdot {\pi }^6-\frac{1}{192}\cdot {\pi }^6$

= $\frac{91}{24}\cdot {\pi }^6$

= ${\left[e^{-3x+4}\right]}_0^3$

$=$ $e^{-3\cdot 3+4}-e^{-3\cdot 0+4}$

= $e^{-9+4}-e^{0+4}$

= $e^{-5}-e^4$

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

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

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

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

= $\frac{1}{2}-\frac{1}{2}\cdot {\pi }^2+\frac{1}{8}\cdot {\pi }^2$

= $\frac{1}{2}-\frac{3}{8}\cdot {\pi }^2$

= ${\left[-\frac{1}{3}{\left(-3x+7\right)}^3-5x\right]}_0^3$

$=$ $-\frac{1}{3}\cdot {\left(-3\cdot 3+7\right)}^3-5\cdot 3-\left(-\frac{1}{3}\cdot {\left(-3\cdot 0+7\right)}^3-5\cdot 0\right)$

= $-\frac{1}{3}\cdot {\left(-9+7\right)}^3-15-\left(-\frac{1}{3}\cdot {\left(0+7\right)}^3+0\right)$

= $-\frac{1}{3}\cdot {\left(-2\right)}^3-15-\left(-\frac{1}{3}\cdot 7^3+0\right)$

= $-\frac{1}{3}\cdot \left(-8\right)-15-\left(-\frac{1}{3}\cdot 343+0\right)$

= $\frac{8}{3}-15-\left(-\frac{343}{3}+0\right)$

= $\frac{8}{3}-\frac{45}{3}-\left(-\frac{343}{3}+0\right)$

= $-\frac{37}{3}+\frac{343}{3}$

= $102$