$$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

Somit gilt:

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

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

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

= $\frac{5}{2}\cdot 9+15-\left(\frac{5}{2}\cdot 1-5\right)$

= $\frac{45}{2}+15-\left(\frac{5}{2}-5\right)$

= $\frac{45}{2}+\frac{30}{2}-\left(\frac{5}{2}-\frac{10}{2}\right)$

= $\frac{75}{2}-1·\left(-\frac{5}{2}\right)$

= $\frac{75}{2}+\frac{5}{2}$

= $40$

= ${\left[\frac{5}{6}{e}^{-3x}+\frac{7}{5}{x}^{\frac{5}{2}}\right]}_0^4$

= ${\left[\frac{5}{6}{e}^{-3x}+\frac{7}{5}{\left(\sqrt{x}\right)}^5\right]}_0^4$

$=$ $\frac{5}{6}{e}^{-3\cdot 4}+\frac{7}{5}{\left(\sqrt{4}\right)}^5-\left(\frac{5}{6}{e}^{-3\cdot 0}+\frac{7}{5}{\left(\sqrt{0}\right)}^5\right)$

= $\frac{5}{6}{e}^{-12}+\frac{7}{5}\cdot 2^5-\left(\frac{5}{6}{e}^0+\frac{7}{5}\cdot 0^5\right)$

= $\frac{5}{6}{e}^{-12}+\frac{7}{5}\cdot 32-\left(\frac{5}{6}+\frac{7}{5}\cdot 0\right)$

= $\frac{5}{6}{e}^{-12}+\frac{224}{5}-\left(\frac{5}{6}+0\right)$

= $\frac{5}{6}{e}^{-12}+\frac{224}{5}-\left(\frac{5}{6}+0\right)$

= $\frac{5}{6}{e}^{-12}+\frac{224}{5}-\frac{5}{6}$

= $\frac{5}{6}{e}^{-12}+\frac{1319}{30}$

= ${\left[-\frac{1}{2}{e}^{-2x+3}\right]}_1^2$

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

= $-\frac{1}{2}{e}^{-4+3}+\frac{1}{2}{e}^{-2+3}$

= $-\frac{1}{2}{e}^{-1}+\frac{1}{2}e$

= $-\frac{1}{2}{e}^{-1}+\frac{1}{2}e$

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

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

= $2\cdot \left(-1\right)+2\cdot 0-\left(2\cdot 0+2\cdot 1\right)$

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

= $-2-2$

= $-4$

= ${\left[-\frac{4}{9}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

= ${\left[-\frac{4}{9}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{8}{3}}^{\frac{13}{3}}$

$=$ $-\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{13}{3}\right)-4}\right)}^3+\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{8}{3}\right)-4}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{13-4}\right)}^3+\frac{4}{9}{\left(\sqrt{8-4}\right)}^3$

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

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

= $-\frac{4}{9}\cdot 27+\frac{4}{9}\cdot 8$

= $-12+\frac{32}{9}$

= $-\frac{108}{9}+\frac{32}{9}$

= $-\frac{76}{9}$