$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ 3 = 3 ⋅ 3 = 9.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{3\; +\; <mn>4</mn>}}{2}$ = 2 ⋅ 3.5 = 7.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 3 +9 +7 = 19

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

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

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

= $-2-8-\left(-\frac{1}{2}+4\right)$

= $-10-\left(-\frac{1}{2}+\frac{8}{2}\right)$

= $-10-1·\frac{7}{2}$

= $-10-\frac{7}{2}$

= $-\frac{20}{2}-\frac{7}{2}$

= $-\frac{27}{2}$

= ${\left[\frac{1}{6}{x}^4-\frac{2}{3}{x}^{\frac{5}{2}}\right]}_0^1$

= ${\left[\frac{1}{6}{x}^4-\frac{2}{3}{\left(\sqrt{x}\right)}^5\right]}_0^1$

$=$ $\frac{1}{6}\cdot 1^4-\frac{2}{3}{\left(\sqrt{1}\right)}^5-\left(\frac{1}{6}\cdot 0^4-\frac{2}{3}{\left(\sqrt{0}\right)}^5\right)$

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

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

= $\frac{1}{6}-\frac{2}{3}-\left(0+0\right)$

= $\frac{1}{6}-\frac{4}{6}+0$

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

= ${\left[-\frac{4}{3}{\left(-x+1\right)}^{\frac{3}{2}}\right]}_{-15}^{-8}$

= ${\left[-\frac{4}{3}{\left(\sqrt{-x+1}\right)}^3\right]}_{-15}^{-8}$

$=$ $-\frac{4}{3}{\left(\sqrt{-\left(-8\right)+1}\right)}^3+\frac{4}{3}{\left(\sqrt{-\left(-15\right)+1}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{8+1}\right)}^3+\frac{4}{3}{\left(\sqrt{15+1}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{9}\right)}^3+\frac{4}{3}{\left(\sqrt{16}\right)}^3$

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

= $-\frac{4}{3}\cdot 27+\frac{4}{3}\cdot 64$

= $-36+\frac{256}{3}$

= $-\frac{108}{3}+\frac{256}{3}$

= $\frac{148}{3}$

= ${\left[\frac{8}{3}\cdot \sin \left(x\right)+4\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $\frac{8}{3}\cdot \sin \left(\frac{1}{2}\pi \right)+4\cdot \cos \left(\frac{1}{2}\pi \right)-\left(\frac{8}{3}\cdot \sin \left(0\right)+4\cdot \cos \left(0\right)\right)$

= $\frac{8}{3}\cdot 1+4\cdot 0-\left(\frac{8}{3}\cdot 0+4\cdot 1\right)$

= $\frac{8}{3}+0-\left(0+4\right)$

= $\frac{8}{3}+0-4$

= $\frac{8}{3}-4$

= $\frac{8}{3}-\frac{12}{3}$

= $-\frac{4}{3}$

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

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

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

= $-\frac{1}{3}\cdot 1+\frac{1}{3}\cdot 0$

= $-\frac{1}{3}+0$

= $-\frac{1}{3}+0$

= $-\frac{1}{3}$