$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -12 -6 +4 +8 = -6

= ${\left[\frac{1}{3}{x}^3-\frac{3}{2}{x}^2\right]}_1^4$

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

= $\frac{1}{3}\cdot 64-\frac{3}{2}\cdot 16-\left(\frac{1}{3}\cdot 1-\frac{3}{2}\cdot 1\right)$

= $\frac{64}{3}-24-\left(\frac{1}{3}-\frac{3}{2}\right)$

= $\frac{64}{3}-\frac{72}{3}-\left(\frac{2}{6}-\frac{9}{6}\right)$

= $-\frac{8}{3}-1·\left(-\frac{7}{6}\right)$

= $-\frac{8}{3}+\frac{7}{6}$

= $-\frac{16}{6}+\frac{7}{6}$

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

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

= ${\left[\frac{1}{x}+\frac{1}{3}{\left(\sqrt{x}\right)}^3\right]}_4^{16}$

$=$ $\frac{1}{16}+\frac{1}{3}{\left(\sqrt{16}\right)}^3-\left(\frac{1}{4}+\frac{1}{3}{\left(\sqrt{4}\right)}^3\right)$

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

= $\frac{1}{16}+\frac{1}{3}\cdot 64-\left(\frac{1}{4}+\frac{1}{3}\cdot 8\right)$

= $\frac{1}{16}+\frac{64}{3}-\left(\frac{1}{4}+\frac{8}{3}\right)$

= $\frac{3}{48}+\frac{1024}{48}-\left(\frac{3}{12}+\frac{32}{12}\right)$

= $\frac{1027}{48}-1·\frac{35}{12}$

= $\frac{1027}{48}-\frac{35}{12}$

= $\frac{1027}{48}-\frac{140}{48}$

= $\frac{1027}{48}-\frac{35}{12}$

= $\frac{1027}{48}-\frac{140}{48}$

= $\frac{887}{48}$

= ${\left[-\frac{2}{9}{\left(-3x+6\right)}^3+x^2\right]}_1^2$

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

= $-\frac{2}{9}\cdot {\left(-6+6\right)}^3+4-\left(-\frac{2}{9}\cdot {\left(-3+6\right)}^3+1\right)$

= $-\frac{2}{9}\cdot 0^3+4-\left(-\frac{2}{9}\cdot 3^3+1\right)$

= $-\frac{2}{9}\cdot 0+4-\left(-\frac{2}{9}\cdot 27+1\right)$

= $0+4-\left(-6+1\right)$

= $4-1·\left(-5\right)$

= $4+5$

= $9$

= ${\left[-x^3+\frac{4}{3}{e}^{-3x}\right]}_{-1}^2$

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

= $-8+\frac{4}{3}{e}^{-6}-\left(-\left(-1\right)+\frac{4}{3}{e}^3\right)$

= $-8+\frac{4}{3}{e}^{-6}-\left(1+\frac{4}{3}{e}^3\right)$

= $\frac{4}{3}{e}^{-6}-8-\left(\frac{4}{3}{e}^3+1\right)$

= $-\frac{4}{3}{e}^3-1·1+\frac{4}{3}{e}^{-6}-8$

= $-\frac{4}{3}{e}^3-1+\frac{4}{3}{e}^{-6}-8$

= $-\frac{4}{3}{e}^3+\frac{4}{3}{e}^{-6}-1-8$

= $-\frac{4}{3}{e}^3+\frac{4}{3}{e}^{-6}-9$

= ${\left[\frac{4}{9}{\left(-3x+4\right)}^{\frac{3}{2}}\right]}_{-7}^{-\frac{5}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{-3x+4}\right)}^3\right]}_{-7}^{-\frac{5}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{-3\cdot \left(-\frac{5}{3}\right)+4}\right)}^3-\frac{4}{9}{\left(\sqrt{-3\cdot \left(-7\right)+4}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{5+4}\right)}^3-\frac{4}{9}{\left(\sqrt{21+4}\right)}^3$

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

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

= $\frac{4}{9}\cdot 27-\frac{4}{9}\cdot 125$

= $12-\frac{500}{9}$

= $\frac{108}{9}-\frac{500}{9}$

= $-\frac{392}{9}$