$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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:

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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 9 +4.5 -1.5 = 12

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

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

= $\frac{3}{2}\cdot 4+10-\left(\frac{3}{2}\cdot 1+5\right)$

= $6+10-\left(\frac{3}{2}+5\right)$

= $16-\left(\frac{3}{2}+\frac{10}{2}\right)$

= $16-1·\frac{13}{2}$

= $16-\frac{13}{2}$

= $\frac{32}{2}-\frac{13}{2}$

= $\frac{19}{2}$

= ${\left[-2\cdot \sin \left(x\right)+\frac{7}{4}{x}^{-1}\right]}_{\pi }^{\frac{3}{2}\pi }$

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

$=$ $-2\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{7}{4\cdot \frac{3}{2}\pi }-\left(-2\cdot \sin \left(\pi \right)+\frac{7}{4\pi }\right)$

= $-2\cdot \left(-1\right)+\frac{7}{4\cdot \frac{3}{2}\pi }-\left(-2\cdot 0+\frac{7}{4\pi }\right)$

= $2+\frac{7}{4\cdot \frac{3}{2}\pi }-\left(0+\frac{7}{4\pi }\right)$

= $2+\frac{7}{6\pi }-\frac{7}{4\pi }$

= $2-\frac{7}{12\pi }$

= ${\left[\frac{2}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[\frac{2}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $\frac{2}{3}{\left(\sqrt{27-2}\right)}^3-\frac{2}{3}{\left(\sqrt{18-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{16}\right)}^3$

= $\frac{2}{3}\cdot 5^3-\frac{2}{3}\cdot 4^3$

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 64$

= $\frac{250}{3}-\frac{128}{3}$

= $\frac{122}{3}$

= ${\left[\frac{7}{6}{e}^{-3x}-\frac{8}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{7}{6}{e}^{-3\cdot \pi }-\frac{8}{3}\cdot \cos \left(\pi \right)-\left(\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}-\frac{8}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }-\frac{8}{3}\cdot \left(-1\right)-\left(\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}-\frac{8}{3}\cdot 0\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{8}{3}-\left(\frac{7}{6}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}+0\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{8}{3}-\frac{7}{6}{e}^{-\frac{3}{2}\pi }$

= ${\left[-\frac{1}{4}{\left(-2x+3\right)}^4-\frac{3}{2}{x}^2\right]}_0^2$

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

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

= $-\frac{1}{4}\cdot {\left(-1\right)}^4-6-\left(-\frac{1}{4}\cdot 3^4+0\right)$

= $-\frac{1}{4}\cdot 1-6-\left(-\frac{1}{4}\cdot 81+0\right)$

= $-\frac{1}{4}-6-\left(-\frac{81}{4}+0\right)$

= $-\frac{1}{4}-\frac{24}{4}-\left(-\frac{81}{4}+0\right)$

= $-\frac{25}{4}+\frac{81}{4}$

= $14$