$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

Somit gilt:

$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = 4.5 = 4.5

= ${\left[-2x^2-5x\right]}_0^4$

$=$ $-2\cdot 4^2-5\cdot 4-\left(-2\cdot 0^2-5\cdot 0\right)$

= $-2\cdot 16-20-\left(-2\cdot 0+0\right)$

= $-32-20-\left(0+0\right)$

= $-52+0$

= $-52$

= ${\left[-\frac{2}{9}{x}^6+2\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{2}{9}\cdot {\pi }^6+2\cdot \sin \left(\pi \right)-\left(-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6+2\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{2}{9}\cdot {\pi }^6+2\cdot 0-\left(-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6+2\cdot 1\right)$

= $-\frac{2}{9}\cdot {\pi }^6+0-\left(-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6+2\right)$

= $-\frac{2}{9}\cdot {\pi }^6-\left(2-\frac{1}{288}\cdot {\pi }^6\right)$

= $-1·2-1·\left(-\frac{1}{288}\cdot {\pi }^6\right)-\frac{2}{9}\cdot {\pi }^6$

= $-2+\frac{1}{288}\cdot {\pi }^6-\frac{2}{9}\cdot {\pi }^6$

= $-2-\frac{7}{32}\cdot {\pi }^6$

= ${\left[-\frac{1}{8}{\left(-2x+1\right)}^4+\frac{5}{2}{x}^2\right]}_1^3$

$=$ $-\frac{1}{8}\cdot {\left(-2\cdot 3+1\right)}^4+\frac{5}{2}\cdot 3^2-\left(-\frac{1}{8}\cdot {\left(-2\cdot 1+1\right)}^4+\frac{5}{2}\cdot 1^2\right)$

= $-\frac{1}{8}\cdot {\left(-6+1\right)}^4+\frac{5}{2}\cdot 9-\left(-\frac{1}{8}\cdot {\left(-2+1\right)}^4+\frac{5}{2}\cdot 1\right)$

= $-\frac{1}{8}\cdot {\left(-5\right)}^4+\frac{45}{2}-\left(-\frac{1}{8}\cdot {\left(-1\right)}^4+\frac{5}{2}\right)$

= $-\frac{1}{8}\cdot 625+\frac{45}{2}-\left(-\frac{1}{8}\cdot 1+\frac{5}{2}\right)$

= $-\frac{625}{8}+\frac{45}{2}-\left(-\frac{1}{8}+\frac{5}{2}\right)$

= $-\frac{625}{8}+\frac{180}{8}-\left(-\frac{1}{8}+\frac{20}{8}\right)$

= $-\frac{445}{8}-1·\frac{19}{8}$

= $-\frac{445}{8}-\frac{19}{8}$

= $-58$

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

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

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

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

= $2+\frac{243}{160}\cdot {\pi }^5-\left(-2+\frac{1}{160}\cdot {\pi }^5\right)$

= $2+\frac{243}{160}\cdot {\pi }^5-1·\left(-2\right)-1·\frac{1}{160}\cdot {\pi }^5$

= $2+\frac{243}{160}\cdot {\pi }^5+2-\frac{1}{160}\cdot {\pi }^5$

= $2+2+\frac{243}{160}\cdot {\pi }^5-\frac{1}{160}\cdot {\pi }^5$

= $4+\frac{121}{80}\cdot {\pi }^5$

= ${\left[e^{3x-7}\right]}_2^5$

$=$ $e^{3\cdot 5-7}-e^{3\cdot 2-7}$

= $e^{15-7}-e^{6-7}$

= $e^8-e^{-1}$