$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 1 = 3 ⋅ 1 = 3.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{1\; +\; <mn>4</mn>}}{2}$ = 2 ⋅ 2.5 = 5.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 1.5 +3 +5 = 9.5

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

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

= $-\frac{5}{3}\cdot 1+1-\left(-\frac{5}{3}\cdot \left(-8\right)-2\right)$

= $-\frac{5}{3}+1-\left(\frac{40}{3}-2\right)$

= $-\frac{5}{3}+\frac{3}{3}-\left(\frac{40}{3}-\frac{6}{3}\right)$

= $-\frac{2}{3}-1·\frac{34}{3}$

= $-\frac{2}{3}-\frac{34}{3}$

= $-12$

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

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

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

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

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

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

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

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

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

= ${\left[-{\left(2x-2\right)}^{\frac{3}{2}}\right]}_3^{\frac{11}{2}}$

= ${\left[-{\left(\sqrt{2x-2}\right)}^3\right]}_3^{\frac{11}{2}}$

$=$ $-{\left(\sqrt{2\cdot \left(\frac{11}{2}\right)-2}\right)}^3+{\left(\sqrt{2\cdot 3-2}\right)}^3$

= $-{\left(\sqrt{11-2}\right)}^3+{\left(\sqrt{6-2}\right)}^3$

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

= $-3^3+2^3$

= $-27+8$

= $-19$

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

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

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

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

= $-2-\frac{9}{8}\cdot {\pi }^3+0$

= $-2-\frac{9}{8}\cdot {\pi }^3$

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

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

= $\frac{1}{3}\cdot 0^3-\frac{1}{3}\cdot 1^3$

= $\frac{1}{3}\cdot 0-\frac{1}{3}\cdot 1$

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

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

= $-\frac{1}{3}$