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

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

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 4 = 2 ⋅ 4 = 8.

Somit gilt:

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

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

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

= $-\frac{1}{3}\cdot 8-\frac{5}{2}\cdot 4-\left(-\frac{1}{3}\cdot \left(-8\right)-\frac{5}{2}\cdot 4\right)$

= $-\frac{8}{3}-10-\left(\frac{8}{3}-10\right)$

= $-\frac{8}{3}-\frac{30}{3}-\left(\frac{8}{3}-\frac{30}{3}\right)$

= $-\frac{38}{3}-1·\left(-\frac{22}{3}\right)$

= $-\frac{38}{3}+\frac{22}{3}$

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

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

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

= $-\frac{2}{3}{e}^{15}-\frac{4}{5}\cdot 3125-\left(-\frac{2}{3}{e}^3-\frac{4}{5}\cdot 1\right)$

= $-\frac{2}{3}{e}^{15}-2500-\left(-\frac{2}{3}{e}^3-\frac{4}{5}\right)$

= $-\frac{2}{3}{e}^{15}-2500+\frac{2}{3}{e}^3-1·\left(-\frac{4}{5}\right)$

= $-\frac{2}{3}{e}^{15}-2500+\frac{2}{3}{e}^3+\frac{4}{5}$

= $-\frac{2}{3}{e}^{15}+\frac{2}{3}{e}^3-2500+\frac{4}{5}$

= $-\frac{2}{3}{e}^{15}+\frac{2}{3}{e}^3-\frac{12496}{5}$

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

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

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

= $\frac{3}{2}\cdot 0-\frac{3}{2}\cdot 0$

= $0+0$

= 0

= ${\left[-8\cdot \cos \left(x\right)+\frac{2}{3}\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $-8\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{2}{3}\cdot \sin \left(\frac{1}{2}\pi \right)-\left(-8\cdot \cos \left(0\right)+\frac{2}{3}\cdot \sin \left(0\right)\right)$

= $-8\cdot 0+\frac{2}{3}\cdot 1-\left(-8\cdot 1+\frac{2}{3}\cdot 0\right)$

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

= $0+\frac{2}{3}+8$

= $\frac{2}{3}+8$

= $\frac{2}{3}+\frac{24}{3}$

= $\frac{26}{3}$

= ${\left[{\left(2x-4\right)}^{\frac{3}{2}}\right]}_{10}^{\frac{29}{2}}$

= ${\left[{\left(\sqrt{2x-4}\right)}^3\right]}_{10}^{\frac{29}{2}}$

$=$ ${\left(\sqrt{2\cdot \left(\frac{29}{2}\right)-4}\right)}^3-{\left(\sqrt{2\cdot 10-4}\right)}^3$

= ${\left(\sqrt{29-4}\right)}^3-{\left(\sqrt{20-4}\right)}^3$

= ${\left(\sqrt{25}\right)}^3-{\left(\sqrt{16}\right)}^3$

= $5^3-4^3$

= $125-64$

= $61$