$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

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

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

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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -6 -3 +4 +8 = 3

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

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

= $\frac{2}{3}\cdot 27+3-\left(\frac{2}{3}\cdot 1+1\right)$

= $18+3-\left(\frac{2}{3}+1\right)$

= $21-\left(\frac{2}{3}+\frac{3}{3}\right)$

= $21-1·\frac{5}{3}$

= $21-\frac{5}{3}$

= $\frac{63}{3}-\frac{5}{3}$

= $\frac{58}{3}$

= ${\left[\frac{9}{4}{e}^{2x}+\frac{9}{8}{x}^4\right]}_{-2}^{-1}$

$=$ $\frac{9}{4}{e}^{2\cdot \left(-1\right)}+\frac{9}{8}\cdot {\left(-1\right)}^4-\left(\frac{9}{4}{e}^{2\cdot \left(-2\right)}+\frac{9}{8}\cdot {\left(-2\right)}^4\right)$

= $\frac{9}{4}{e}^{-2}+\frac{9}{8}\cdot 1-\left(\frac{9}{4}{e}^{-4}+\frac{9}{8}\cdot 16\right)$

= $\frac{9}{4}{e}^{-2}+\frac{9}{8}-\left(\frac{9}{4}{e}^{-4}+18\right)$

= $\frac{9}{4}{e}^{-2}+\frac{9}{8}-\frac{9}{4}{e}^{-4}-1·18$

= $\frac{9}{4}{e}^{-2}+\frac{9}{8}-\frac{9}{4}{e}^{-4}-18$

= $\frac{9}{4}{e}^{-2}-\frac{9}{4}{e}^{-4}+\frac{9}{8}-18$

= $\frac{9}{4}{e}^{-2}-\frac{9}{4}{e}^{-4}-\frac{135}{8}$

= ${\left[e^{-3x+4}\right]}_0^2$

$=$ $e^{-3\cdot 2+4}-e^{-3\cdot 0+4}$

= $e^{-6+4}-e^{0+4}$

= $e^{-2}-e^4$

= ${\left[-\frac{2}{3}{x}^3+\frac{10}{9}{x}^{\frac{3}{2}}\right]}_1^4$

= ${\left[-\frac{2}{3}{x}^3+\frac{10}{9}{\left(\sqrt{x}\right)}^3\right]}_1^4$

$=$ $-\frac{2}{3}\cdot 4^3+\frac{10}{9}{\left(\sqrt{4}\right)}^3-\left(-\frac{2}{3}\cdot 1^3+\frac{10}{9}{\left(\sqrt{1}\right)}^3\right)$

= $-\frac{2}{3}\cdot 64+\frac{10}{9}\cdot 2^3-\left(-\frac{2}{3}\cdot 1+\frac{10}{9}\cdot 1^3\right)$

= $-\frac{128}{3}+\frac{10}{9}\cdot 8-\left(-\frac{2}{3}+\frac{10}{9}\cdot 1\right)$

= $-\frac{128}{3}+\frac{80}{9}-\left(-\frac{2}{3}+\frac{10}{9}\right)$

= $-\frac{384}{9}+\frac{80}{9}-\left(-\frac{6}{9}+\frac{10}{9}\right)$

= $-\frac{304}{9}-1·\frac{4}{9}$

= $-\frac{304}{9}-\frac{4}{9}$

= $-\frac{308}{9}$

= ${\left[\frac{1}{9}{\left(3x-6\right)}^3+3x^2\right]}_2^3$

$=$ $\frac{1}{9}\cdot {\left(3\cdot 3-6\right)}^3+3\cdot 3^2-\left(\frac{1}{9}\cdot {\left(3\cdot 2-6\right)}^3+3\cdot 2^2\right)$

= $\frac{1}{9}\cdot {\left(9-6\right)}^3+3\cdot 9-\left(\frac{1}{9}\cdot {\left(6-6\right)}^3+3\cdot 4\right)$

= $\frac{1}{9}\cdot 3^3+27-\left(\frac{1}{9}\cdot 0^3+12\right)$

= $\frac{1}{9}\cdot 27+27-\left(\frac{1}{9}\cdot 0+12\right)$

= $3+27-\left(0+12\right)$

= $30-12$

= $18$