$$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

Somit gilt:

$$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = -3 = -3

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

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

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

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

= $-\frac{8}{3}+\frac{6}{3}-\left(\frac{1}{3}-\frac{3}{3}\right)$

= $-\frac{2}{3}-1·\left(-\frac{2}{3}\right)$

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

= 0

= ${\left[\frac{7}{2}{e}^{2x}-\frac{9}{8}{x}^{-2}\right]}_2^3$

= ${\left[\frac{7}{2}{e}^{2x}-\frac{9}{8x^2}\right]}_2^3$

$=$ $\frac{7}{2}{e}^{2\cdot 3}-\frac{9}{8\cdot 3^2}-\left(\frac{7}{2}{e}^{2\cdot 2}-\frac{9}{8\cdot 2^2}\right)$

= $\frac{7}{2}{e}^6-\frac{9}{8}\cdot \left(\frac{1}{9}\right)-\left(\frac{7}{2}{e}^4-\frac{9}{8}\cdot \left(\frac{1}{4}\right)\right)$

= $\frac{7}{2}{e}^6-\frac{1}{8}-\left(\frac{7}{2}{e}^4-\frac{9}{32}\right)$

= $\frac{7}{2}{e}^6-\frac{1}{8}-\frac{7}{2}{e}^4-1·\left(-\frac{9}{32}\right)$

= $\frac{7}{2}{e}^6-\frac{1}{8}-\frac{7}{2}{e}^4+\frac{9}{32}$

= $\frac{7}{2}{e}^6-\frac{7}{2}{e}^4-\frac{1}{8}+\frac{9}{32}$

= $\frac{7}{2}{e}^6-\frac{7}{2}{e}^4+\frac{5}{32}$

= ${\left[-e^{x-1}\right]}_1^3$

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

= $-e^2+e^0$

= $-e^2+1$

= ${\left[\frac{3}{4}\cdot \sin \left(x\right)-2\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $\frac{3}{4}\cdot \sin \left(\frac{1}{2}\pi \right)-2\cdot \cos \left(\frac{1}{2}\pi \right)-\left(\frac{3}{4}\cdot \sin \left(0\right)-2\cdot \cos \left(0\right)\right)$

= $\frac{3}{4}\cdot 1-2\cdot 0-\left(\frac{3}{4}\cdot 0-2\cdot 1\right)$

= $\frac{3}{4}+0-\left(0-2\right)$

= $\frac{3}{4}+0+2$

= $\frac{3}{4}+2$

= $\frac{3}{4}+\frac{8}{4}$

= $\frac{11}{4}$

= ${\left[\frac{2}{9}{\left(3x-6\right)}^3-2x^2\right]}_0^1$

$=$ $\frac{2}{9}\cdot {\left(3\cdot 1-6\right)}^3-2\cdot 1^2-\left(\frac{2}{9}\cdot {\left(3\cdot 0-6\right)}^3-2\cdot 0^2\right)$

= $\frac{2}{9}\cdot {\left(3-6\right)}^3-2\cdot 1-\left(\frac{2}{9}\cdot {\left(0-6\right)}^3-2\cdot 0\right)$

= $\frac{2}{9}\cdot {\left(-3\right)}^3-2-\left(\frac{2}{9}\cdot {\left(-6\right)}^3+0\right)$

= $\frac{2}{9}\cdot \left(-27\right)-2-\left(\frac{2}{9}\cdot \left(-216\right)+0\right)$

= $-6-2-\left(-48+0\right)$

= $-8+48$

= $40$