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

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 -3 = 0

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

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

= $-2\cdot 4-10-\left(-2\cdot 1+5\right)$

= $-8-10-\left(-2+5\right)$

= $-18-1·3$

= $-18-3$

= $-21$

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

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

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

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

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

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

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

= $\frac{5}{2}-\frac{1}{2{\pi }^2}+\frac{5}{2}+\frac{2}{{\pi }^2}$

= $\frac{5}{2}+\frac{5}{2}-\frac{1}{2{\pi }^2}+\frac{2}{{\pi }^2}$

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

= ${\left[\frac{1}{2}{e}^{2x-1}\right]}_2^4$

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

= $\frac{1}{2}{e}^{8-1}-\frac{1}{2}{e}^{4-1}$

= $\frac{1}{2}{e}^7-\frac{1}{2}{e}^3$

= ${\left[\frac{7}{4}\cdot \cos \left(x\right)+\frac{1}{2}{e}^x\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{7}{4}\cdot \cos \left(\pi \right)+\frac{1}{2}{e}^{\pi }-\left(\frac{7}{4}\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{1}{2}{e}^{\frac{1}{2}\pi }\right)$

= $\frac{7}{4}\cdot \left(-1\right)+\frac{1}{2}{e}^{\pi }-\left(\frac{7}{4}\cdot 0+\frac{1}{2}{e}^{\frac{1}{2}\pi }\right)$

= $-\frac{7}{4}+\frac{1}{2}{e}^{\pi }-\left(0+\frac{1}{2}{e}^{\frac{1}{2}\pi }\right)$

= $\frac{1}{2}{e}^{\pi }-\frac{7}{4}-\frac{1}{2}{e}^{\frac{1}{2}\pi }$

= $\frac{1}{2}{e}^{\pi }-\frac{1}{2}{e}^{\frac{1}{2}\pi }-\frac{7}{4}$

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

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

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

= $-0+0$

= 0