$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

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.

Somit gilt:

$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = -9 -4.5 = -13.5

= ${\left[-\frac{5}{3}{x}^3-4x\right]}_{-2}^{-1}$

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

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

= $\frac{5}{3}+4-\left(\frac{40}{3}+8\right)$

= $\frac{5}{3}+\frac{12}{3}-\left(\frac{40}{3}+\frac{24}{3}\right)$

= $\frac{17}{3}-1·\frac{64}{3}$

= $\frac{17}{3}-\frac{64}{3}$

= $-\frac{47}{3}$

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

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

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

= $\frac{1}{5}\cdot 243-\frac{5}{2}\cdot \left(\frac{1}{9}\right)-\left(\frac{1}{5}\cdot 1-\frac{5}{2}\cdot 1\right)$

= $\frac{243}{5}-\frac{5}{18}-\left(\frac{1}{5}-\frac{5}{2}\right)$

= $\frac{4374}{90}-\frac{25}{90}-\left(\frac{2}{10}-\frac{25}{10}\right)$

= $\frac{4349}{90}-1·\left(-\frac{23}{10}\right)$

= $\frac{4349}{90}+\frac{23}{10}$

= $\frac{4349}{90}+\frac{207}{90}$

= $\frac{2278}{45}$

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

$=$ $-\frac{1}{3}{e}^{-3\cdot 3+3}+\frac{1}{3}{e}^{-3\cdot 1+3}$

= $-\frac{1}{3}{e}^{-9+3}+\frac{1}{3}{e}^{-3+3}$

= $-\frac{1}{3}{e}^{-6}+\frac{1}{3}{e}^0$

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

= ${\left[3\cdot \cos \left(x\right)-\frac{5}{6}{e}^{-3x}\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $3\cdot \left(-1\right)-\frac{5}{6}{e}^{-3\cdot \pi }-\left(3\cdot 0-\frac{5}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-3-\frac{5}{6}{e}^{-3\cdot \pi }-\left(0-\frac{5}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{5}{6}{e}^{-3\cdot \pi }-3+\frac{5}{6}{e}^{-\frac{3}{2}\pi }$

= ${\left[\frac{1}{6}{\left(-2x+1\right)}^3+3x\right]}_2^5$

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

= $\frac{1}{6}\cdot {\left(-10+1\right)}^3+15-\left(\frac{1}{6}\cdot {\left(-4+1\right)}^3+6\right)$

= $\frac{1}{6}\cdot {\left(-9\right)}^3+15-\left(\frac{1}{6}\cdot {\left(-3\right)}^3+6\right)$

= $\frac{1}{6}\cdot \left(-729\right)+15-\left(\frac{1}{6}\cdot \left(-27\right)+6\right)$

= $-\frac{243}{2}+15-\left(-\frac{9}{2}+6\right)$

= $-\frac{243}{2}+\frac{30}{2}-\left(-\frac{9}{2}+\frac{12}{2}\right)$

= $-\frac{213}{2}-1·\frac{3}{2}$

= $-\frac{213}{2}-\frac{3}{2}$

= $-108$