$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

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

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 4 = 2 ⋅ 4 = 8.

Somit gilt:

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

= ${\left[\frac{5}{3}{x}^3-2x\right]}_0^3$

$=$ $\frac{5}{3}\cdot 3^3-2\cdot 3-\left(\frac{5}{3}\cdot 0^3-2\cdot 0\right)$

= $\frac{5}{3}\cdot 27-6-\left(\frac{5}{3}\cdot 0+0\right)$

= $45-6-\left(0+0\right)$

= $39+0$

= $39$

= ${\left[\frac{2}{3}{x}^{-1}-\frac{4}{5}{x}^{\frac{5}{4}}\right]}_4^{16}$

= ${\left[\frac{2}{3x}-\frac{4}{5}{\left(\sqrt[4]{x}\right)}^5\right]}_4^{16}$

$=$ $\frac{2}{3\cdot 16}-\frac{4}{5}{\left(\sqrt[4]{16}\right)}^5-\left(\frac{2}{3\cdot 4}-\frac{4}{5}{\left(\sqrt[4]{4}\right)}^5\right)$

= $\frac{2}{3}\cdot \left(\frac{1}{16}\right)-\frac{4}{5}\cdot 2^5-\left(\frac{2}{3}\cdot \left(\frac{1}{4}\right)-\frac{4}{5}\cdot 1^5\right)$

= $\frac{1}{24}-\frac{4}{5}\cdot 32-\left(\frac{1}{6}-\frac{4}{5}\cdot 1\right)$

= $\frac{1}{24}-\frac{128}{5}-\left(\frac{1}{6}-\frac{4}{5}\right)$

= $\frac{5}{120}-\frac{3072}{120}-\left(\frac{5}{30}-\frac{24}{30}\right)$

= $-\frac{3067}{120}-1·\left(-\frac{19}{30}\right)$

= $-\frac{3067}{120}+\frac{19}{30}$

= $-\frac{3067}{120}+\frac{76}{120}$

= $-\frac{3067}{120}+\frac{19}{30}$

= $-\frac{3067}{120}+\frac{76}{120}$

= $-\frac{997}{40}$

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

= ${\left[-\frac{2}{3}{\left(\sqrt{-3x+4}\right)}^3\right]}_{-4}^{-\frac{5}{3}}$

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

= $-\frac{2}{3}{\left(\sqrt{5+4}\right)}^3+\frac{2}{3}{\left(\sqrt{12+4}\right)}^3$

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

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

= $-\frac{2}{3}\cdot 27+\frac{2}{3}\cdot 64$

= $-18+\frac{128}{3}$

= $-\frac{54}{3}+\frac{128}{3}$

= $\frac{74}{3}$

= ${\left[-9\cdot \sin \left(x\right)+\frac{3}{4}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-9\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{3}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(-9\cdot \sin \left(\frac{1}{2}\pi \right)+\frac{3}{4}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

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

= $9+0-\left(-9+0\right)$

= $9+9$

= $18$

= ${\left[\frac{2}{3}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{3}}^{\frac{28}{3}}$

= ${\left[\frac{2}{3}{\left(\sqrt{3x-3}\right)}^3\right]}_{\frac{7}{3}}^{\frac{28}{3}}$

$=$ $\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{28}{3}\right)-3}\right)}^3-\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{28-3}\right)}^3-\frac{2}{3}{\left(\sqrt{7-3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{4}\right)}^3$

= $\frac{2}{3}\cdot 5^3-\frac{2}{3}\cdot 2^3$

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 8$

= $\frac{250}{3}-\frac{16}{3}$

= $78$