$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

Somit gilt:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 4.5 -4.5 = 0

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

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

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

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

= $6-\left(\frac{1}{2}-\frac{4}{2}\right)$

= $6-1·\left(-\frac{3}{2}\right)$

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

= $\frac{12}{2}+\frac{3}{2}$

= $\frac{15}{2}$

= ${\left[\frac{7}{4}\cdot \cos \left(x\right)-\frac{3}{4}\cdot \sin \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $\frac{7}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\frac{3}{4}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(\frac{7}{4}\cdot \cos \left(0\right)-\frac{3}{4}\cdot \sin \left(0\right)\right)$

= $\frac{7}{4}\cdot 0-\frac{3}{4}\cdot \left(-1\right)-\left(\frac{7}{4}\cdot 1-\frac{3}{4}\cdot 0\right)$

= $0+\frac{3}{4}-\left(\frac{7}{4}+0\right)$

= $0+\frac{3}{4}-\left(\frac{7}{4}+0\right)$

= $\frac{3}{4}-\frac{7}{4}$

= $-1$

= ${\left[\frac{2}{9}{\left(3x-7\right)}^3+x^2\right]}_2^4$

$=$ $\frac{2}{9}\cdot {\left(3\cdot 4-7\right)}^3+4^2-\left(\frac{2}{9}\cdot {\left(3\cdot 2-7\right)}^3+2^2\right)$

= $\frac{2}{9}\cdot {\left(12-7\right)}^3+16-\left(\frac{2}{9}\cdot {\left(6-7\right)}^3+4\right)$

= $\frac{2}{9}\cdot 5^3+16-\left(\frac{2}{9}\cdot {\left(-1\right)}^3+4\right)$

= $\frac{2}{9}\cdot 125+16-\left(\frac{2}{9}\cdot \left(-1\right)+4\right)$

= $\frac{250}{9}+16-\left(-\frac{2}{9}+4\right)$

= $\frac{250}{9}+\frac{144}{9}-\left(-\frac{2}{9}+\frac{36}{9}\right)$

= $\frac{394}{9}-1·\frac{34}{9}$

= $\frac{394}{9}-\frac{34}{9}$

= $40$

= ${\left[-\frac{9}{4}\cdot \cos \left(x\right)+\frac{3}{2}{x}^{-3}\right]}_{\pi }^{2\pi }$

= ${\left[-\frac{9}{4}\cdot \cos \left(x\right)+\frac{3}{2x^3}\right]}_{\pi }^{2\pi }$

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

= $-\frac{9}{4}\cdot 1+\frac{3}{2{\left(2\pi \right)}^3}-\left(-\frac{9}{4}\cdot \left(-1\right)+\frac{3}{2{\pi }^3}\right)$

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

= $-\frac{9}{4}+\frac{3}{16{\pi }^3}-\left(\frac{9}{4}+\frac{3}{2{\pi }^3}\right)$

= $-\frac{9}{4}+\frac{3}{16{\pi }^3}-1·\frac{9}{4}-1·\frac{3}{2{\pi }^3}$

= $-\frac{9}{4}+\frac{3}{16{\pi }^3}-\frac{9}{4}-\frac{3}{2{\pi }^3}$

= $-\frac{9}{4}-\frac{9}{4}+\frac{3}{16{\pi }^3}-\frac{3}{2{\pi }^3}$

= $-\frac{9}{2}-\frac{21}{16{\pi }^3}$

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

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

= $-\cos \left(-\frac{7}{2}\pi \right)+\cos \left(-\frac{5}{2}\pi \right)$

= $-0+0$

= 0