$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

Somit gilt:

$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 4.5 -3 = 1.5

= ${\left[-\frac{5}{2}{x}^2+2x\right]}_1^4$

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

= $-\frac{5}{2}\cdot 16+8-\left(-\frac{5}{2}\cdot 1+2\right)$

= $-40+8-\left(-\frac{5}{2}+2\right)$

= $-32-\left(-\frac{5}{2}+\frac{4}{2}\right)$

= $-32-1·\left(-\frac{1}{2}\right)$

= $-32+\frac{1}{2}$

= $-\frac{64}{2}+\frac{1}{2}$

= $-\frac{63}{2}$

= ${\left[-\frac{4}{3}{x}^{\frac{3}{2}}+\frac{1}{2}{x}^4\right]}_0^9$

= ${\left[-\frac{4}{3}{\left(\sqrt{x}\right)}^3+\frac{1}{2}{x}^4\right]}_0^9$

$=$ $-\frac{4}{3}{\left(\sqrt{9}\right)}^3+\frac{1}{2}\cdot 9^4-\left(-\frac{4}{3}{\left(\sqrt{0}\right)}^3+\frac{1}{2}\cdot 0^4\right)$

= $-\frac{4}{3}\cdot 3^3+\frac{1}{2}\cdot 6561-\left(-\frac{4}{3}\cdot 0^3+\frac{1}{2}\cdot 0\right)$

= $-\frac{4}{3}\cdot 27+\frac{6561}{2}-\left(-\frac{4}{3}\cdot 0+0\right)$

= $-36+\frac{6561}{2}-\left(0+0\right)$

= $-\frac{72}{2}+\frac{6561}{2}+0$

= $\frac{6489}{2}$

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

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

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

= $-3\cdot 1+3\cdot 0$

= $-3+0$

= $-3$

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

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

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

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

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

= $3-\frac{10}{3\pi }-\left(-3-\frac{10}{\pi }\right)$

= $3-\frac{10}{3\pi }-1·\left(-3\right)-1·\left(-\frac{10}{\pi }\right)$

= $3-\frac{10}{3\pi }+3+\frac{10}{\pi }$

= $3+3-\frac{10}{3\pi }+\frac{10}{\pi }$

= $6+\frac{20}{3\pi }$

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

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

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

= $-\frac{1}{2}\cdot 0+\frac{1}{2}\cdot 0$

= $0+0$

= 0