$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ gibt des orientierten (also mit Vorzeichen behafteten) Flächeninhalt zwischen dem Graph der Funktion und der x-Achse an.

Um diesen aus dem Schaubild zu berechnen unterteilen wir die Fläche in Rechtecke, Parallelogrammen und ggf. Dreiecke:

I₁ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

Somit gilt:

$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = 4.5 -4 = 0.5

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

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

= $-\frac{1}{2}\cdot 4-8-\left(-\frac{1}{2}\cdot 9-12\right)$

= $-2-8-\left(-\frac{9}{2}-12\right)$

= $-10-\left(-\mathrm{4,5}-12\right)$

= $-10-1·\left(-\mathrm{16,5}\right)$

= $-10+\mathrm{16,5}$

= $\mathrm{6,5}$

= ${\left[-\frac{1}{4}\cdot \sin \left(x\right)+5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

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

= $\frac{1}{4}+\frac{1}{4}$

= $\frac{1}{2}$

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

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

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

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

= $0+1$

= $1$

= ${\left[8x^{\frac{1}{2}}-\frac{3}{2}\cdot \sin \left(x\right)\right]}_1^9$

= ${\left[8\sqrt{x}-\frac{3}{2}\cdot \sin \left(x\right)\right]}_1^9$

$=$ $8\sqrt{9}-\frac{3}{2}\cdot \sin \left(9\right)-\left(8\sqrt{1}-\frac{3}{2}\cdot \sin \left(1\right)\right)$

= $8\cdot 3-\frac{3}{2}\cdot \sin \left(9\right)-\left(8\cdot 1-\frac{3}{2}\cdot \sin \left(1\right)\right)$

= $24-\frac{3}{2}\cdot \sin \left(9\right)-\left(8-\frac{3}{2}\cdot \sin \left(1\right)\right)$

= $-\frac{3}{2}\cdot \sin \left(9\right)+24-\left(-\frac{3}{2}\cdot \sin \left(1\right)+8\right)$

= $-\frac{3}{2}\cdot \sin \left(9\right)+24+\frac{3}{2}\cdot \sin \left(1\right)-8$

= $-\frac{3}{2}\cdot \sin \left(9\right)+\frac{3}{2}\cdot \sin \left(1\right)+24-8$

= $-\frac{3}{2}\cdot \sin \left(9\right)+\frac{3}{2}\cdot \sin \left(1\right)+16$

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

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

= $\frac{2}{3}\cdot \sin \left(-\frac{11}{2}\pi \right)-\frac{2}{3}\cdot \sin \left(-\frac{5}{2}\pi \right)$

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

= $\frac{2}{3}+\frac{2}{3}$

= $\frac{4}{3}$