$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = 1 +2 = 3

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

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

= $\frac{2}{3}\cdot \left(-1\right)+2-\left(\frac{2}{3}\cdot \left(-27\right)+6\right)$

= $-\frac{2}{3}+2-\left(-18+6\right)$

= $-\frac{2}{3}+\frac{6}{3}-1·\left(-12\right)$

= $\frac{4}{3}+12$

= $\frac{4}{3}+\frac{36}{3}$

= $\frac{40}{3}$

= ${\left[-2x^{\frac{5}{4}}+\frac{3}{2}{e}^{2x}\right]}_4^{16}$

= ${\left[-2{\left(\sqrt[4]{x}\right)}^5+\frac{3}{2}{e}^{2x}\right]}_4^{16}$

$=$ $-2{\left(\sqrt[4]{16}\right)}^5+\frac{3}{2}{e}^{2\cdot 16}-\left(-2{\left(\sqrt[4]{4}\right)}^5+\frac{3}{2}{e}^{2\cdot 4}\right)$

= $-2\cdot 2^5+\frac{3}{2}{e}^{32}-\left(-2\cdot 1^5+\frac{3}{2}{e}^8\right)$

= $-2\cdot 32+\frac{3}{2}{e}^{32}-\left(-2\cdot 1+\frac{3}{2}{e}^8\right)$

= $-64+\frac{3}{2}{e}^{32}-\left(-2+\frac{3}{2}{e}^8\right)$

= $\frac{3}{2}{e}^{32}-64-\left(\frac{3}{2}{e}^8-2\right)$

= $\frac{3}{2}{e}^{32}-64-\frac{3}{2}{e}^8-1·\left(-2\right)$

= $\frac{3}{2}{e}^{32}-64-\frac{3}{2}{e}^8+2$

= $\frac{3}{2}{e}^{32}-\frac{3}{2}{e}^8-64+2$

= $\frac{3}{2}{e}^{32}-\frac{3}{2}{e}^8-62$

= ${\left[\frac{1}{3}{\left(-2x+1\right)}^3\right]}_1^4$

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

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

= $\frac{1}{3}\cdot {\left(-7\right)}^3-\frac{1}{3}\cdot {\left(-1\right)}^3$

= $\frac{1}{3}\cdot \left(-343\right)-\frac{1}{3}\cdot \left(-1\right)$

= $-\frac{343}{3}+\frac{1}{3}$

= $-114$

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

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

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

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

= $-1-1$

= $-2$

= ${\left[\frac{1}{3}\ln \left(\left|$ 3x-4$\right|\right)\right]}_2^4$

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

= $\frac{1}{3}\ln \left(\left|$ 12-4$\right|\right)-\frac{1}{3}\ln \left(\left|$ 6-4$\right|\right)$

= $\frac{1}{3}\ln \left(8\right)-\frac{1}{3}\ln \left(\left|$ 6-4$\right|\right)$

= $\frac{1}{3}\ln \left(8\right)-\frac{1}{3}\ln \left(2\right)$