$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

Somit gilt:

$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 2 -6 = -4

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

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

= $\frac{5}{3}\cdot 8-8-\left(\frac{5}{3}\cdot \left(-1\right)+4\right)$

= $\frac{40}{3}-8-\left(-\frac{5}{3}+4\right)$

= $\frac{40}{3}-\frac{24}{3}-\left(-\frac{5}{3}+\frac{12}{3}\right)$

= $\frac{16}{3}-1·\frac{7}{3}$

= $\frac{16}{3}-\frac{7}{3}$

= $3$

= ${\left[-\frac{1}{4}{x}^6-\frac{1}{2}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{1}{4}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{1}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{1}{4}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{1}{2}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

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

= $-\frac{1}{4}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{1}{2}-\left(-\frac{1}{4}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{1}{2}\right)$

= $\frac{1}{2}-\frac{729}{256}\cdot {\pi }^6-\left(-\frac{1}{2}-\frac{1}{256}\cdot {\pi }^6\right)$

= $\frac{1}{2}-\frac{729}{256}\cdot {\pi }^6-1·\left(-\frac{1}{2}\right)-1·\left(-\frac{1}{256}\cdot {\pi }^6\right)$

= $\frac{1}{2}-\frac{729}{256}\cdot {\pi }^6+\frac{1}{2}+\frac{1}{256}\cdot {\pi }^6$

= $\frac{1}{2}+\frac{1}{2}-\frac{729}{256}\cdot {\pi }^6+\frac{1}{256}\cdot {\pi }^6$

= $1-\frac{91}{32}\cdot {\pi }^6$

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

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

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

= $-\frac{1}{4}\cdot 1^4+\frac{1}{4}\cdot 4^4$

= $-\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 256$

= $-\frac{1}{4}+64$

= $-\frac{1}{4}+\frac{256}{4}$

= $\frac{255}{4}$

= ${\left[-\frac{9}{2}\cdot \cos \left(x\right)-2\cdot \sin \left(x\right)\right]}_0^{\pi }$

$=$ $-\frac{9}{2}\cdot \cos \left(\pi \right)-2\cdot \sin \left(\pi \right)-\left(-\frac{9}{2}\cdot \cos \left(0\right)-2\cdot \sin \left(0\right)\right)$

= $-\frac{9}{2}\cdot \left(-1\right)-2\cdot 0-\left(-\frac{9}{2}\cdot 1-2\cdot 0\right)$

= $\frac{9}{2}+0-\left(-\frac{9}{2}+0\right)$

= $\frac{9}{2}+0-\left(-\frac{9}{2}+0\right)$

= $\frac{9}{2}+\frac{9}{2}$

= $9$

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

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

= $\sin \left(-\frac{7}{2}\pi \right)-\sin \left(-2\pi \right)$

= $1-0$

= $1$