$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

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

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

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

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

= $-18+\frac{9}{2}-\left(\frac{2}{3}+\frac{1}{2}\right)$

= $-\frac{36}{2}+\frac{9}{2}-\left(\frac{4}{6}+\frac{3}{6}\right)$

= $-\frac{27}{2}-1·\frac{7}{6}$

= $-\frac{27}{2}-\frac{7}{6}$

= $-\frac{81}{6}-\frac{7}{6}$

= $-\frac{44}{3}$

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

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

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

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

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

= $\frac{16}{81{\pi }^3}-\left(2+\frac{2}{3{\pi }^3}\right)$

= $-1·2-1·\frac{2}{3{\pi }^3}+\frac{16}{81{\pi }^3}$

= $-2-\frac{2}{3{\pi }^3}+\frac{16}{81{\pi }^3}$

= $-2-\frac{38}{81{\pi }^3}$

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

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

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

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

= $\frac{1}{6}\cdot \left(-8\right)-\frac{1}{6}\cdot 64$

= $-\frac{4}{3}-\frac{32}{3}$

= $-12$

= ${\left[\frac{10}{3}{x}^{\frac{3}{2}}+\frac{2}{3}{x}^3\right]}_4^{16}$

= ${\left[\frac{10}{3}{\left(\sqrt{x}\right)}^3+\frac{2}{3}{x}^3\right]}_4^{16}$

$=$ $\frac{10}{3}{\left(\sqrt{16}\right)}^3+\frac{2}{3}\cdot {16}^3-\left(\frac{10}{3}{\left(\sqrt{4}\right)}^3+\frac{2}{3}\cdot 4^3\right)$

= $\frac{10}{3}\cdot 4^3+\frac{2}{3}\cdot 4096-\left(\frac{10}{3}\cdot 2^3+\frac{2}{3}\cdot 64\right)$

= $\frac{10}{3}\cdot 64+\frac{8192}{3}-\left(\frac{10}{3}\cdot 8+\frac{128}{3}\right)$

= $\frac{640}{3}+\frac{8192}{3}-\left(\frac{80}{3}+\frac{128}{3}\right)$

= $2944-1·\frac{208}{3}$

= $2944-\frac{208}{3}$

= $\frac{8832}{3}-\frac{208}{3}$

= $\frac{8624}{3}$

= ${\left[-\frac{1}{3}{e}^{-3x+5}\right]}_2^4$

$=$ $-\frac{1}{3}{e}^{-3\cdot 4+5}+\frac{1}{3}{e}^{-3\cdot 2+5}$

= $-\frac{1}{3}{e}^{-12+5}+\frac{1}{3}{e}^{-6+5}$

= $-\frac{1}{3}{e}^{-7}+\frac{1}{3}{e}^{-1}$