$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ 1 = 3 ⋅ 1 = 3.

Somit gilt:

$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 1 +3 = 4

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

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

= $\frac{2}{3}\cdot \left(-1\right)+2\cdot 1-\left(\frac{2}{3}\cdot \left(-27\right)+2\cdot 9\right)$

= $-\frac{2}{3}+2-\left(-18+18\right)$

= $-\frac{2}{3}+\frac{6}{3}-1·0$

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

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

= $\frac{4}{3}$

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

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

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

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

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

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }-4+\frac{1}{3}{e}^{\frac{3}{2}\pi }-1·4$

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }-4+\frac{1}{3}{e}^{\frac{3}{2}\pi }-4$

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }+\frac{1}{3}{e}^{\frac{3}{2}\pi }-4-4$

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }+\frac{1}{3}{e}^{\frac{3}{2}\pi }-8$

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

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

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

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

= $0+0$

= 0

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

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

= $-4\cdot 0+4\cdot \left(-1\right)-\left(-4\cdot 1+4\cdot 0\right)$

= $0-4-\left(-4+0\right)$

= $-4+4$

= 0

= ${\left[-\frac{1}{3}{\left(2x-2\right)}^{\frac{3}{2}}\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

= ${\left[-\frac{1}{3}{\left(\sqrt{2x-2}\right)}^3\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

$=$ $-\frac{1}{3}{\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3+\frac{1}{3}{\left(\sqrt{2\cdot \left(\frac{11}{2}\right)-2}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{27-2}\right)}^3+\frac{1}{3}{\left(\sqrt{11-2}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{25}\right)}^3+\frac{1}{3}{\left(\sqrt{9}\right)}^3$

= $-\frac{1}{3}\cdot 5^3+\frac{1}{3}\cdot 3^3$

= $-\frac{1}{3}\cdot 125+\frac{1}{3}\cdot 27$

= $-\frac{125}{3}+9$

= $-\frac{125}{3}+\frac{27}{3}$

= $-\frac{98}{3}$