$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 4 = 3 ⋅ 4 = 12.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{4\; +\; <mn>2</mn>}}{2}$ = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 6 +12 +6 = 24

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

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

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

= $-\frac{32}{3}-6-\left(\frac{4}{3}-\frac{3}{2}\right)$

= $-\frac{32}{3}-\frac{18}{3}-\left(\frac{8}{6}-\frac{9}{6}\right)$

= $-\frac{50}{3}-1·\left(-\frac{1}{6}\right)$

= $-\frac{50}{3}+\frac{1}{6}$

= $-\frac{100}{6}+\frac{1}{6}$

= $-\frac{33}{2}$

= ${\left[-4x^{\frac{3}{2}}-\frac{1}{6}{x}^6\right]}_1^9$

= ${\left[-4{\left(\sqrt{x}\right)}^3-\frac{1}{6}{x}^6\right]}_1^9$

$=$ $-4{\left(\sqrt{9}\right)}^3-\frac{1}{6}\cdot 9^6-\left(-4{\left(\sqrt{1}\right)}^3-\frac{1}{6}\cdot 1^6\right)$

= $-4\cdot 3^3-\frac{1}{6}\cdot 531441-\left(-4\cdot 1^3-\frac{1}{6}\cdot 1\right)$

= $-4\cdot 27-\frac{177147}{2}-\left(-4\cdot 1-\frac{1}{6}\right)$

= $-108-\frac{177147}{2}-\left(-4-\frac{1}{6}\right)$

= $-\frac{216}{2}-\frac{177147}{2}-\left(-\frac{24}{6}-\frac{1}{6}\right)$

= $-\frac{177363}{2}-1·\left(-\frac{25}{6}\right)$

= $-\frac{177363}{2}+\frac{25}{6}$

= $-\frac{532089}{6}+\frac{25}{6}$

= $-\frac{177363}{2}+\frac{25}{6}$

= $-\frac{532089}{6}+\frac{25}{6}$

= $-\frac{266032}{3}$

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

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

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

= $-3\cdot 0+3\cdot 1$

= $0+3$

= $3$

= ${\left[3\cdot \sin \left(x\right)-4\cdot \cos \left(x\right)\right]}_0^{\pi }$

$=$ $3\cdot \sin \left(\pi \right)-4\cdot \cos \left(\pi \right)-\left(3\cdot \sin \left(0\right)-4\cdot \cos \left(0\right)\right)$

= $3\cdot 0-4\cdot \left(-1\right)-\left(3\cdot 0-4\cdot 1\right)$

= $0+4-\left(0-4\right)$

= $4+4$

= $8$

= ${\left[-\frac{1}{3}{\left(x-2\right)}^3-\frac{5}{2}{x}^2\right]}_1^4$

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

= $-\frac{1}{3}\cdot 2^3-\frac{5}{2}\cdot 16-\left(-\frac{1}{3}\cdot {\left(-1\right)}^3-\frac{5}{2}\cdot 1\right)$

= $-\frac{1}{3}\cdot 8-40-\left(-\frac{1}{3}\cdot \left(-1\right)-\frac{5}{2}\right)$

= $-\frac{8}{3}-40-\left(\frac{1}{3}-\frac{5}{2}\right)$

= $-\frac{8}{3}-\frac{120}{3}-\left(\frac{2}{6}-\frac{15}{6}\right)$

= $-\frac{128}{3}-1·\left(-\frac{13}{6}\right)$

= $-\frac{128}{3}+\frac{13}{6}$

= $-\frac{256}{6}+\frac{13}{6}$

= $-\frac{128}{3}+\frac{13}{6}$

= $-\frac{256}{6}+\frac{13}{6}$

= $-\frac{81}{2}$