$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 6 -3 -9 = -6

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

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

= $0+0-\left(9+6\right)$

= $-1·9-1·6$

= $-9-6$

= $-15$

= ${\left[-\frac{2}{3}{x}^3+\frac{10}{3}{x}^{\frac{3}{2}}\right]}_0^1$

= ${\left[-\frac{2}{3}{x}^3+\frac{10}{3}{\left(\sqrt{x}\right)}^3\right]}_0^1$

$=$ $-\frac{2}{3}\cdot 1^3+\frac{10}{3}{\left(\sqrt{1}\right)}^3-\left(-\frac{2}{3}\cdot 0^3+\frac{10}{3}{\left(\sqrt{0}\right)}^3\right)$

= $-\frac{2}{3}\cdot 1+\frac{10}{3}\cdot 1^3-\left(-\frac{2}{3}\cdot 0+\frac{10}{3}\cdot 0^3\right)$

= $-\frac{2}{3}+\frac{10}{3}\cdot 1-\left(0+\frac{10}{3}\cdot 0\right)$

= $-\frac{2}{3}+\frac{10}{3}-\left(0+0\right)$

= $\frac{8}{3}+0$

= $\frac{8}{3}$

= ${\left[-\frac{4}{9}{\left(-3x+7\right)}^{\frac{3}{2}}\right]}_{-6}^{-3}$

= ${\left[-\frac{4}{9}{\left(\sqrt{-3x+7}\right)}^3\right]}_{-6}^{-3}$

$=$ $-\frac{4}{9}{\left(\sqrt{-3\cdot \left(-3\right)+7}\right)}^3+\frac{4}{9}{\left(\sqrt{-3\cdot \left(-6\right)+7}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{9+7}\right)}^3+\frac{4}{9}{\left(\sqrt{18+7}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{16}\right)}^3+\frac{4}{9}{\left(\sqrt{25}\right)}^3$

= $-\frac{4}{9}\cdot 4^3+\frac{4}{9}\cdot 5^3$

= $-\frac{4}{9}\cdot 64+\frac{4}{9}\cdot 125$

= $-\frac{256}{9}+\frac{500}{9}$

= $\frac{244}{9}$

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

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

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

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

= $-3\cdot \sin \left(16\right)+\frac{2}{3}\cdot 64-\left(-3\cdot \sin \left(4\right)+\frac{2}{3}\cdot 8\right)$

= $-3\cdot \sin \left(16\right)+\frac{128}{3}-\left(-3\cdot \sin \left(4\right)+\frac{16}{3}\right)$

= $-3\cdot \sin \left(16\right)+\frac{128}{3}-1·\left(-3\cdot \sin \left(4\right)\right)-1·\frac{16}{3}$

= $-3\cdot \sin \left(16\right)+\frac{128}{3}+3\cdot \sin \left(4\right)-\frac{16}{3}$

= $-3\cdot \sin \left(16\right)+3\cdot \sin \left(4\right)+\frac{128}{3}-\frac{16}{3}$

= $-3\cdot \sin \left(16\right)+3\cdot \sin \left(4\right)+\frac{112}{3}$

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

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

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

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

= $\frac{1}{2}+\frac{1}{2}$

= $1$