$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 4 = 2 ⋅ 4 = 8.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 -3 +6 +8 = 7

= ${\left[\frac{4}{3}{x}^3+x\right]}_{-2}^0$

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

= $\frac{4}{3}\cdot 0+0-\left(\frac{4}{3}\cdot \left(-8\right)-2\right)$

= $0+0-\left(-\frac{32}{3}-2\right)$

= $0-\left(-\frac{32}{3}-\frac{6}{3}\right)$

= $-1·\left(-\frac{38}{3}\right)$

= $\frac{38}{3}$

= ${\left[-2x^{\frac{3}{2}}-2x^{-1}\right]}_4^{16}$

= ${\left[-2{\left(\sqrt{x}\right)}^3-\frac{2}{x}\right]}_4^{16}$

$=$ $-2{\left(\sqrt{16}\right)}^3-\frac{2}{16}-\left(-2{\left(\sqrt{4}\right)}^3-\frac{2}{4}\right)$

= $-2\cdot 4^3-2\cdot \left(\frac{1}{16}\right)-\left(-2\cdot 2^3-2\cdot \left(\frac{1}{4}\right)\right)$

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

= $-128-\frac{1}{8}-\left(-16-\frac{1}{2}\right)$

= $-\frac{1024}{8}-\frac{1}{8}-\left(-\frac{32}{2}-\frac{1}{2}\right)$

= $-\frac{1025}{8}-1·\left(-\frac{33}{2}\right)$

= $-\frac{1025}{8}+\frac{33}{2}$

= $-\frac{1025}{8}+\frac{132}{8}$

= $-\frac{1025}{8}+\frac{33}{2}$

= $-\frac{1025}{8}+\frac{132}{8}$

= $-\frac{893}{8}$

= ${\left[-\frac{2}{9}{\left(3x-6\right)}^{\frac{3}{2}}\right]}_5^{\frac{22}{3}}$

= ${\left[-\frac{2}{9}{\left(\sqrt{3x-6}\right)}^3\right]}_5^{\frac{22}{3}}$

$=$ $-\frac{2}{9}{\left(\sqrt{3\cdot \left(\frac{22}{3}\right)-6}\right)}^3+\frac{2}{9}{\left(\sqrt{3\cdot 5-6}\right)}^3$

= $-\frac{2}{9}{\left(\sqrt{22-6}\right)}^3+\frac{2}{9}{\left(\sqrt{15-6}\right)}^3$

= $-\frac{2}{9}{\left(\sqrt{16}\right)}^3+\frac{2}{9}{\left(\sqrt{9}\right)}^3$

= $-\frac{2}{9}\cdot 4^3+\frac{2}{9}\cdot 3^3$

= $-\frac{2}{9}\cdot 64+\frac{2}{9}\cdot 27$

= $-\frac{128}{9}+6$

= $-\frac{128}{9}+\frac{54}{9}$

= $-\frac{74}{9}$

= ${\left[-9\cdot \cos \left(x\right)+2x^{-2}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

= ${\left[-9\cdot \cos \left(x\right)+\frac{2}{x^2}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-9\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{2}{{\left(\frac{3}{2}\pi \right)}^2}-\left(-9\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{2}{{\left(\frac{1}{2}\pi \right)}^2}\right)$

= $-9\cdot 0+\frac{2}{{\left(\frac{3}{2}\pi \right)}^2}-\left(-9\cdot 0+\frac{2}{{\left(\frac{1}{2}\pi \right)}^2}\right)$

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

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

= $\frac{8}{9{\pi }^2}-\frac{72}{9{\pi }^2}$

= $-\frac{64}{9{\pi }^2}$

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

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

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

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

= $0+3$

= $3$