$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

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.

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:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 -3 +3 = -6

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

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

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

= $-\frac{5}{2}-3-\left(-10+6\right)$

= $-\frac{5}{2}-\frac{6}{2}-1·\left(-4\right)$

= $-\frac{11}{2}+4$

= $-\frac{11}{2}+\frac{8}{2}$

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

= ${\left[\frac{9}{2}\cdot \sin \left(x\right)-\frac{7}{6}{e}^{-3x}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{9}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\frac{7}{6}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(\frac{9}{2}\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $\frac{9}{2}\cdot \left(-1\right)-\frac{7}{6}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(\frac{9}{2}\cdot 1-\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{9}{2}-\frac{7}{6}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(\frac{9}{2}-\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{7}{6}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\frac{9}{2}-\left(-\frac{7}{6}{e}^{-\frac{3}{2}\pi }+\frac{9}{2}\right)$

= $-\frac{7}{6}{e}^{-\frac{9}{2}\pi }-\frac{9}{2}+\frac{7}{6}{e}^{-\frac{3}{2}\pi }-1·\frac{9}{2}$

= $-\frac{7}{6}{e}^{-\frac{9}{2}\pi }-\frac{9}{2}+\frac{7}{6}{e}^{-\frac{3}{2}\pi }-\frac{9}{2}$

= $\frac{7}{6}{e}^{-\frac{3}{2}\pi }-\frac{7}{6}{e}^{-\frac{9}{2}\pi }-\frac{9}{2}-\frac{9}{2}$

= $\frac{7}{6}{e}^{-\frac{3}{2}\pi }-\frac{7}{6}{e}^{-\frac{9}{2}\pi }-9$

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

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

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

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

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

= $\frac{125}{6}+\frac{1}{6}$

= $21$

= ${\left[\frac{1}{4}{x}^4-3x^{\frac{3}{2}}\right]}_0^9$

= ${\left[\frac{1}{4}{x}^4-3{\left(\sqrt{x}\right)}^3\right]}_0^9$

$=$ $\frac{1}{4}\cdot 9^4-3{\left(\sqrt{9}\right)}^3-\left(\frac{1}{4}\cdot 0^4-3{\left(\sqrt{0}\right)}^3\right)$

= $\frac{1}{4}\cdot 6561-3\cdot 3^3-\left(\frac{1}{4}\cdot 0-3\cdot 0^3\right)$

= $\frac{6561}{4}-3\cdot 27-\left(0-3\cdot 0\right)$

= $\frac{6561}{4}-81-\left(0+0\right)$

= $\frac{6561}{4}-\frac{324}{4}+0$

= $\frac{6237}{4}$

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

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

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

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

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

= $-\frac{1}{3}-\frac{9}{3}+2$

= $-\frac{10}{3}+2$

= $-\frac{10}{3}+\frac{6}{3}$

= $-\frac{4}{3}$