$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 1 = 2 ⋅ 1 = 2.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{1\; +\; <mn>2</mn>}}{2}$ = 2 ⋅ 1.5 = 3.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 1 +2 +3 = 6

= ${\left[-x^3-3x\right]}_0^1$

$=$ $-1^3-3\cdot 1-\left(-0^3-3\cdot 0\right)$

= $-1-3-\left(-0+0\right)$

= $-1-3$

= $-4$

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

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

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

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

= $-\frac{1}{2}{e}^{3\cdot \pi }+\frac{7}{2}+\frac{1}{2}{e}^{\frac{3}{2}\pi }$

= $-\frac{1}{2}{e}^{3\pi }+\frac{1}{2}{e}^{\frac{3}{2}\pi }+\frac{7}{2}$

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

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

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

= $-\frac{3}{2}\cdot 0+\frac{3}{2}\cdot 0$

= $0+0$

= 0

= ${\left[\frac{4}{3}{x}^3+\frac{5}{4}\cdot \cos \left(x\right)\right]}_0^{\pi }$

$=$ $\frac{4}{3}\cdot {\pi }^3+\frac{5}{4}\cdot \cos \left(\pi \right)-\left(\frac{4}{3}\cdot {\left(0\right)}^3+\frac{5}{4}\cdot \cos \left(0\right)\right)$

= $\frac{4}{3}\cdot {\pi }^3+\frac{5}{4}\cdot \left(-1\right)-\left(\frac{4}{3}\cdot 0+\frac{5}{4}\cdot 1\right)$

= $\frac{4}{3}\cdot {\pi }^3-\frac{5}{4}-\left(0+\frac{5}{4}\right)$

= $-\frac{5}{4}+\frac{4}{3}\cdot {\pi }^3-\left(0+\frac{5}{4}\right)$

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

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

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

= ${\left[\frac{2}{3}{\left(-x+3\right)}^{\frac{3}{2}}\right]}_{-22}^{-13}$

= ${\left[\frac{2}{3}{\left(\sqrt{-x+3}\right)}^3\right]}_{-22}^{-13}$

$=$ $\frac{2}{3}{\left(\sqrt{-\left(-13\right)+3}\right)}^3-\frac{2}{3}{\left(\sqrt{-\left(-22\right)+3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{13+3}\right)}^3-\frac{2}{3}{\left(\sqrt{22+3}\right)}^3$

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

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

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

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

= $-\frac{122}{3}$