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

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ 3 = 3 ⋅ 3 = 9.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4.5 +9 = 13.5

= ${\left[\frac{1}{2}{x}^2-3x\right]}_{-1}^0$

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

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

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

= $0-\left(\frac{1}{2}+\frac{6}{2}\right)$

= $-1·\frac{7}{2}$

= $-\frac{7}{2}$

= ${\left[-\frac{5}{24}{x}^6-\cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{24}\cdot {\pi }^6-\cos \left(\pi \right)-\left(-\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6-\cos \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{24}\cdot {\pi }^6-\left(-1\right)-\left(-\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6-0\right)$

= $-\frac{5}{24}\cdot {\pi }^6+1-\left(-\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6+0\right)$

= $1-\frac{5}{24}\cdot {\pi }^6+\frac{5}{1536}\cdot {\pi }^6$

= $1-\frac{105}{512}\cdot {\pi }^6$

= ${\left[-3\cdot \cos \left(-x+\frac{1}{2}\pi \right)\right]}_0^{\pi }$

$=$ $-3\cdot \cos \left(-\pi +\frac{1}{2}\pi \right)+3\cdot \cos \left(-\left(0\right)+\frac{1}{2}\pi \right)$

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

= $-3\cdot 0+3\cdot 0$

= $0+0$

= 0

= ${\left[-4x^{-1}-\frac{7}{9}{e}^{-3x}\right]}_2^6$

= ${\left[-\frac{4}{x}-\frac{7}{9}{e}^{-3x}\right]}_2^6$

$=$ $-\frac{4}{6}-\frac{7}{9}{e}^{-3\cdot 6}-\left(-\frac{4}{2}-\frac{7}{9}{e}^{-3\cdot 2}\right)$

= $-4\cdot \left(\frac{1}{6}\right)-\frac{7}{9}{e}^{-18}-\left(-4\cdot \left(\frac{1}{2}\right)-\frac{7}{9}{e}^{-6}\right)$

= $-\frac{2}{3}-\frac{7}{9}{e}^{-18}-\left(-2-\frac{7}{9}{e}^{-6}\right)$

= $-\frac{7}{9}{e}^{-18}-\frac{2}{3}-\left(-\frac{7}{9}{e}^{-6}-2\right)$

= $\frac{7}{9}{e}^{-6}-1·\left(-2\right)-\frac{7}{9}{e}^{-18}-\frac{2}{3}$

= $\frac{7}{9}{e}^{-6}+2-\frac{7}{9}{e}^{-18}-\frac{2}{3}$

= $\frac{7}{9}{e}^{-6}-\frac{7}{9}{e}^{-18}+2-\frac{2}{3}$

= $\frac{7}{9}{e}^{-6}-\frac{7}{9}{e}^{-18}+\frac{4}{3}$

= ${\left[\frac{1}{8}{\left(2x-1\right)}^4\right]}_0^2$

$=$ $\frac{1}{8}\cdot {\left(2\cdot 2-1\right)}^4-\frac{1}{8}\cdot {\left(2\cdot 0-1\right)}^4$

= $\frac{1}{8}\cdot {\left(4-1\right)}^4-\frac{1}{8}\cdot {\left(0-1\right)}^4$

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

= $\frac{1}{8}\cdot 81-\frac{1}{8}\cdot 1$

= $\frac{81}{8}-\frac{1}{8}$

= $10$