$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

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

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

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{2\; +\; <mn>3</mn>}}{2}$ = 2 ⋅ 2.5 = 5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4.5 +3 +4 +5 = 7.5

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

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

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

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

= $\frac{40}{3}-\frac{24}{3}-\left(-\frac{5}{3}-\frac{6}{3}\right)$

= $\frac{16}{3}-1·\left(-\frac{11}{3}\right)$

= $\frac{16}{3}+\frac{11}{3}$

= $9$

= ${\left[-\frac{4}{3}{x}^{\frac{3}{2}}+\frac{3}{2}{x}^2\right]}_4^9$

= ${\left[-\frac{4}{3}{\left(\sqrt{x}\right)}^3+\frac{3}{2}{x}^2\right]}_4^9$

$=$ $-\frac{4}{3}{\left(\sqrt{9}\right)}^3+\frac{3}{2}\cdot 9^2-\left(-\frac{4}{3}{\left(\sqrt{4}\right)}^3+\frac{3}{2}\cdot 4^2\right)$

= $-\frac{4}{3}\cdot 3^3+\frac{3}{2}\cdot 81-\left(-\frac{4}{3}\cdot 2^3+\frac{3}{2}\cdot 16\right)$

= $-\frac{4}{3}\cdot 27+\frac{243}{2}-\left(-\frac{4}{3}\cdot 8+24\right)$

= $-36+\frac{243}{2}-\left(-\frac{32}{3}+24\right)$

= $-\frac{72}{2}+\frac{243}{2}-\left(-\frac{32}{3}+\frac{72}{3}\right)$

= $\frac{171}{2}-1·\frac{40}{3}$

= $\frac{171}{2}-\frac{40}{3}$

= $\frac{513}{6}-\frac{80}{6}$

= $\frac{171}{2}-\frac{40}{3}$

= $\frac{513}{6}-\frac{80}{6}$

= $\frac{433}{6}$

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

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

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

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

= $\frac{1}{3}+0$

= $\frac{1}{3}+0$

= $\frac{1}{3}$

= ${\left[-\frac{1}{2}\cdot \cos \left(x\right)-\frac{2}{5}{x}^5\right]}_0^{\pi }$

$=$ $-\frac{1}{2}\cdot \cos \left(\pi \right)-\frac{2}{5}\cdot {\pi }^5-\left(-\frac{1}{2}\cdot \cos \left(0\right)-\frac{2}{5}\cdot {\left(0\right)}^5\right)$

= $-\frac{1}{2}\cdot \left(-1\right)-\frac{2}{5}\cdot {\pi }^5-\left(-\frac{1}{2}\cdot 1-\frac{2}{5}\cdot 0\right)$

= $\frac{1}{2}-\frac{2}{5}\cdot {\pi }^5-\left(-\frac{1}{2}+0\right)$

= $\frac{1}{2}-\frac{2}{5}\cdot {\pi }^5-\left(-\frac{1}{2}+0\right)$

= $\frac{1}{2}-\frac{2}{5}\cdot {\pi }^5+\frac{1}{2}$

= $\frac{1}{2}+\frac{1}{2}-\frac{2}{5}\cdot {\pi }^5$

= $1-\frac{2}{5}\cdot {\pi }^5$

= ${\left[-\frac{3}{2}{\left(2x-3\right)}^{-1}\right]}_2^3$

= ${\left[-\frac{3}{2\left(2x-3\right)}\right]}_2^3$

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

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

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

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

= $-\frac{1}{2}+\frac{3}{2}$

= $1$