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

Somit gilt:

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

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

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

= $\frac{5}{3}\cdot 1+5-\left(\frac{5}{3}\cdot \left(-27\right)-15\right)$

= $\frac{5}{3}+5-\left(-45-15\right)$

= $\frac{5}{3}+\frac{15}{3}-1·\left(-60\right)$

= $\frac{20}{3}+60$

= $\frac{20}{3}+\frac{180}{3}$

= $\frac{200}{3}$

= ${\left[\frac{3}{8}{e}^{-2x}+\frac{9}{5}{x}^5\right]}_{-3}^1$

$=$ $\frac{3}{8}{e}^{-2\cdot 1}+\frac{9}{5}\cdot 1^5-\left(\frac{3}{8}{e}^{-2\cdot \left(-3\right)}+\frac{9}{5}\cdot {\left(-3\right)}^5\right)$

= $\frac{3}{8}{e}^{-2}+\frac{9}{5}\cdot 1-\left(\frac{3}{8}{e}^6+\frac{9}{5}\cdot \left(-243\right)\right)$

= $\frac{3}{8}{e}^{-2}+\frac{9}{5}-\left(\frac{3}{8}{e}^6-\frac{2187}{5}\right)$

= $-\frac{3}{8}{e}^6-1·\left(-\frac{2187}{5}\right)+\frac{3}{8}{e}^{-2}+\frac{9}{5}$

= $-\frac{3}{8}{e}^6+\frac{2187}{5}+\frac{3}{8}{e}^{-2}+\frac{9}{5}$

= $-\frac{3}{8}{e}^6+\frac{3}{8}{e}^{-2}+\frac{2187}{5}+\frac{9}{5}$

= $-\frac{3}{8}{e}^6+\frac{3}{8}{e}^{-2}+\frac{2196}{5}$

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

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

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

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

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

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

= $\frac{1}{4}+\frac{8}{4}-1·5$

= $\frac{9}{4}-5$

= $\frac{9}{4}-\frac{20}{4}$

= $\frac{9}{4}-5$

= $\frac{9}{4}-\frac{20}{4}$

= $-\frac{11}{4}$

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

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

= $-1+7\cdot 0-\left(0+7\cdot 1\right)$

= $-1+0-\left(0+7\right)$

= $-1-7$

= $-8$

= ${\left[-\frac{1}{2}{\left(2x-1\right)}^3-6x\right]}_2^5$

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

= $-\frac{1}{2}\cdot {\left(10-1\right)}^3-30-\left(-\frac{1}{2}\cdot {\left(4-1\right)}^3-12\right)$

= $-\frac{1}{2}\cdot 9^3-30-\left(-\frac{1}{2}\cdot 3^3-12\right)$

= $-\frac{1}{2}\cdot 729-30-\left(-\frac{1}{2}\cdot 27-12\right)$

= $-\frac{729}{2}-30-\left(-\frac{27}{2}-12\right)$

= $-\frac{729}{2}-\frac{60}{2}-\left(-\frac{27}{2}-\frac{24}{2}\right)$

= $-\frac{789}{2}-1·\left(-\frac{51}{2}\right)$

= $-\frac{789}{2}+\frac{51}{2}$

= $-369$