$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

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

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4.5 +2 +4 = 1.5

= ${\left[\frac{5}{3}{x}^3-4x\right]}_{-1}^1$

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

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

= $\frac{5}{3}-4-\left(-\frac{5}{3}+4\right)$

= $\frac{5}{3}-\frac{12}{3}-\left(-\frac{5}{3}+\frac{12}{3}\right)$

= $-\frac{7}{3}-1·\frac{7}{3}$

= $-\frac{7}{3}-\frac{7}{3}$

= $-\frac{14}{3}$

= ${\left[-2x^{\frac{3}{2}}-\frac{1}{4}{e}^{-3x}\right]}_4^9$

= ${\left[-2{\left(\sqrt{x}\right)}^3-\frac{1}{4}{e}^{-3x}\right]}_4^9$

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

= $-2\cdot 3^3-\frac{1}{4}{e}^{-27}-\left(-2\cdot 2^3-\frac{1}{4}{e}^{-12}\right)$

= $-2\cdot 27-\frac{1}{4}{e}^{-27}-\left(-2\cdot 8-\frac{1}{4}{e}^{-12}\right)$

= $-54-\frac{1}{4}{e}^{-27}-\left(-16-\frac{1}{4}{e}^{-12}\right)$

= $-\frac{1}{4}{e}^{-27}-54-\left(-\frac{1}{4}{e}^{-12}-16\right)$

= $\frac{1}{4}{e}^{-12}-1·\left(-16\right)-\frac{1}{4}{e}^{-27}-54$

= $\frac{1}{4}{e}^{-12}+16-\frac{1}{4}{e}^{-27}-54$

= $\frac{1}{4}{e}^{-12}-\frac{1}{4}{e}^{-27}+16-54$

= $\frac{1}{4}{e}^{-12}-\frac{1}{4}{e}^{-27}-38$

= ${\left[\frac{1}{6}{\left(-2x+5\right)}^3\right]}_1^3$

$=$ $\frac{1}{6}\cdot {\left(-2\cdot 3+5\right)}^3-\frac{1}{6}\cdot {\left(-2\cdot 1+5\right)}^3$

= $\frac{1}{6}\cdot {\left(-6+5\right)}^3-\frac{1}{6}\cdot {\left(-2+5\right)}^3$

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

= $\frac{1}{6}\cdot \left(-1\right)-\frac{1}{6}\cdot 27$

= $-\frac{1}{6}-\frac{9}{2}$

= $-\frac{1}{6}-\frac{27}{6}$

= $-\frac{14}{3}$

= ${\left[-\frac{5}{2}{x}^{-2}-\frac{5}{2}{x}^2\right]}_2^6$

= ${\left[-\frac{5}{2x^2}-\frac{5}{2}{x}^2\right]}_2^6$

$=$ $-\frac{5}{2\cdot 6^2}-\frac{5}{2}\cdot 6^2-\left(-\frac{5}{2\cdot 2^2}-\frac{5}{2}\cdot 2^2\right)$

= $-\frac{5}{2}\cdot \left(\frac{1}{36}\right)-\frac{5}{2}\cdot 36-\left(-\frac{5}{2}\cdot \left(\frac{1}{4}\right)-\frac{5}{2}\cdot 4\right)$

= $-\frac{5}{72}-90-\left(-\frac{5}{8}-10\right)$

= $-\frac{5}{72}-\frac{6480}{72}-\left(-\frac{5}{8}-\frac{80}{8}\right)$

= $-\frac{6485}{72}-1·\left(-\frac{85}{8}\right)$

= $-\frac{6485}{72}+\frac{85}{8}$

= $-\frac{6485}{72}+\frac{765}{72}$

= $-\frac{715}{9}$

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

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

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

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

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

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

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

= $-\frac{15}{4}-1·\left(-\frac{7}{4}\right)$

= $-\frac{15}{4}+\frac{7}{4}$

= $-2$