$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{-2\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4.5 -2 -6 -5 = -8.5

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

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

= $-\frac{4}{3}\cdot 8+6-\left(-\frac{4}{3}\cdot \left(-1\right)-3\right)$

= $-\frac{32}{3}+6-\left(\frac{4}{3}-3\right)$

= $-\frac{32}{3}+\frac{18}{3}-\left(\frac{4}{3}-\frac{9}{3}\right)$

= $-\frac{14}{3}-1·\left(-\frac{5}{3}\right)$

= $-\frac{14}{3}+\frac{5}{3}$

= $-3$

= ${\left[-\frac{7}{6}{x}^{\frac{3}{2}}-\frac{1}{6}{x}^4\right]}_1^4$

= ${\left[-\frac{7}{6}{\left(\sqrt{x}\right)}^3-\frac{1}{6}{x}^4\right]}_1^4$

$=$ $-\frac{7}{6}{\left(\sqrt{4}\right)}^3-\frac{1}{6}\cdot 4^4-\left(-\frac{7}{6}{\left(\sqrt{1}\right)}^3-\frac{1}{6}\cdot 1^4\right)$

= $-\frac{7}{6}\cdot 2^3-\frac{1}{6}\cdot 256-\left(-\frac{7}{6}\cdot 1^3-\frac{1}{6}\cdot 1\right)$

= $-\frac{7}{6}\cdot 8-\frac{128}{3}-\left(-\frac{7}{6}\cdot 1-\frac{1}{6}\right)$

= $-\frac{28}{3}-\frac{128}{3}-\left(-\frac{7}{6}-\frac{1}{6}\right)$

= $-52-1·\left(-\frac{4}{3}\right)$

= $-52+\frac{4}{3}$

= $-\frac{156}{3}+\frac{4}{3}$

= $-\frac{152}{3}$

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

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

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

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

= $\frac{1}{3}+0$

= $\frac{1}{3}+0$

= $\frac{1}{3}$

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

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

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

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

= $-\frac{3}{2}{e}^{-2\pi }-\left(0-\frac{3}{2}\right)$

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

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

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

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

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

= $\frac{2}{3}+2-\left(\frac{16}{3}+1\right)$

= $\frac{2}{3}+\frac{6}{3}-\left(\frac{16}{3}+\frac{3}{3}\right)$

= $\frac{8}{3}-1·\frac{19}{3}$

= $\frac{8}{3}-\frac{19}{3}$

= $-\frac{11}{3}$