$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

Somit gilt:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 6 -1.5 = 4.5

= ${\left[-x^3+\frac{5}{2}{x}^2\right]}_{-1}^2$

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

= $-8+\frac{5}{2}\cdot 4-\left(-\left(-1\right)+\frac{5}{2}\cdot 1\right)$

= $-8+10-\left(1+\frac{5}{2}\right)$

= $2-\left(\frac{2}{2}+\frac{5}{2}\right)$

= $2-1·\frac{7}{2}$

= $2-\frac{7}{2}$

= $\frac{4}{2}-\frac{7}{2}$

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

= ${\left[\frac{1}{2}{x}^4+\frac{1}{2}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{1}{2}\cdot {\pi }^4+\frac{1}{2}\cdot \cos \left(\pi \right)-\left(\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4+\frac{1}{2}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{2}\cdot {\pi }^4+\frac{1}{2}\cdot \left(-1\right)-\left(\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4+\frac{1}{2}\cdot 0\right)$

= $\frac{1}{2}\cdot {\pi }^4-\frac{1}{2}-\left(\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4+0\right)$

= $-\frac{1}{2}+\frac{1}{2}\cdot {\pi }^4-\frac{1}{32}\cdot {\pi }^4$

= $-\frac{1}{2}+\frac{15}{32}\cdot {\pi }^4$

= ${\left[-e^{-2x+3}\right]}_0^2$

$=$ $-e^{-2\cdot 2+3}+e^{-2\cdot 0+3}$

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

= $-e^{-1}+e^3$

= ${\left[-\frac{4}{3}\cdot \cos \left(x\right)+\frac{5}{6}{x}^6\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{4}{3}\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{5}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6-\left(-\frac{4}{3}\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6\right)$

= $-\frac{4}{3}\cdot 0+\frac{5}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6-\left(-\frac{4}{3}\cdot 0+\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6\right)$

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

= $\frac{1215}{128}\cdot {\pi }^6-\frac{5}{384}\cdot {\pi }^6$

= $\frac{3645}{384}\cdot {\pi }^6-\frac{5}{384}\cdot {\pi }^6$

= $\frac{455}{48}\cdot {\pi }^6$

= ${\left[e^{x-1}\right]}_1^4$

$=$ $e^{4-1}-e^{1-1}$

= $e^3-e^0$

= $e^3-1$