$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

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

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4.5 +3 = -1.5

= ${\left[-\frac{3}{2}{x}^2-4x\right]}_{-1}^1$

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

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

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

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

= $-\frac{11}{2}-1·\frac{5}{2}$

= $-\frac{11}{2}-\frac{5}{2}$

= $-8$

= ${\left[-\frac{7}{12}{x}^6-\frac{2}{3}{e}^{2x}\right]}_{-2}^1$

$=$ $-\frac{7}{12}\cdot 1^6-\frac{2}{3}{e}^{2\cdot 1}-\left(-\frac{7}{12}\cdot {\left(-2\right)}^6-\frac{2}{3}{e}^{2\cdot \left(-2\right)}\right)$

= $-\frac{7}{12}\cdot 1-\frac{2}{3}{e}^2-\left(-\frac{7}{12}\cdot 64-\frac{2}{3}{e}^{-4}\right)$

= $-\frac{7}{12}-\frac{2}{3}{e}^2-\left(-\frac{112}{3}-\frac{2}{3}{e}^{-4}\right)$

= $-\frac{2}{3}{e}^2-\frac{7}{12}-\left(-\frac{2}{3}{e}^{-4}-\frac{112}{3}\right)$

= $-\frac{2}{3}{e}^2-\frac{7}{12}+\frac{2}{3}{e}^{-4}-1·\left(-\frac{112}{3}\right)$

= $-\frac{2}{3}{e}^2-\frac{7}{12}+\frac{2}{3}{e}^{-4}+\frac{112}{3}$

= $-\frac{2}{3}{e}^2+\frac{2}{3}{e}^{-4}-\frac{7}{12}+\frac{112}{3}$

= $-\frac{2}{3}{e}^2+\frac{2}{3}{e}^{-4}+\frac{147}{4}$

= ${\left[-\frac{1}{2}{\left(-2x+3\right)}^3+6x\right]}_1^4$

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

= $-\frac{1}{2}\cdot {\left(-8+3\right)}^3+24-\left(-\frac{1}{2}\cdot {\left(-2+3\right)}^3+6\right)$

= $-\frac{1}{2}\cdot {\left(-5\right)}^3+24-\left(-\frac{1}{2}\cdot 1^3+6\right)$

= $-\frac{1}{2}\cdot \left(-125\right)+24-\left(-\frac{1}{2}\cdot 1+6\right)$

= $\frac{125}{2}+24-\left(-\frac{1}{2}+6\right)$

= $\frac{125}{2}+\frac{48}{2}-\left(-\frac{1}{2}+\frac{12}{2}\right)$

= $\frac{173}{2}-1·\frac{11}{2}$

= $\frac{173}{2}-\frac{11}{2}$

= $81$

= ${\left[-\frac{7}{2}\cdot \sin \left(x\right)+\frac{3}{4}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{7}{2}\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{3}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(-\frac{7}{2}\cdot \sin \left(\frac{1}{2}\pi \right)+\frac{3}{4}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{7}{2}\cdot \left(-1\right)+\frac{3}{4}\cdot 0-\left(-\frac{7}{2}\cdot 1+\frac{3}{4}\cdot 0\right)$

= $\frac{7}{2}+0-\left(-\frac{7}{2}+0\right)$

= $\frac{7}{2}+0-\left(-\frac{7}{2}+0\right)$

= $\frac{7}{2}+\frac{7}{2}$

= $7$

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

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

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

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

= $-\frac{1}{2}\cdot 64+\frac{1}{2}\cdot \left(-8\right)$

= $-32-4$

= $-36$