$$\underset{2}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

Somit gilt:

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

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

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

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

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

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

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

= $\frac{7}{3}+\frac{7}{3}$

= $\frac{14}{3}$

= ${\left[-\frac{1}{6}{e}^{-3x}-\frac{1}{5}{x}^{\frac{5}{4}}\right]}_0^4$

= ${\left[-\frac{1}{6}{e}^{-3x}-\frac{1}{5}{\left(\sqrt[4]{x}\right)}^5\right]}_0^4$

$=$ $-\frac{1}{6}{e}^{-3\cdot 4}-\frac{1}{5}{\left(\sqrt[4]{4}\right)}^5-\left(-\frac{1}{6}{e}^{-3\cdot 0}-\frac{1}{5}{\left(\sqrt[4]{0}\right)}^5\right)$

= $-\frac{1}{6}{e}^{-12}-\frac{1}{5}\cdot 1^5-\left(-\frac{1}{6}{e}^0-\frac{1}{5}\cdot 0^5\right)$

= $-\frac{1}{6}{e}^{-12}-\frac{1}{5}\cdot 1-\left(-\frac{1}{6}-\frac{1}{5}\cdot 0\right)$

= $-\frac{1}{6}{e}^{-12}-\frac{1}{5}-\left(-\frac{1}{6}+0\right)$

= $-\frac{1}{6}{e}^{-12}-\frac{1}{5}-\left(-\frac{1}{6}+0\right)$

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

= $-\frac{1}{6}{e}^{-12}-\frac{1}{30}$

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

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

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

= $-\left(-8\right)+1$

= $8+1$

= $9$

= ${\left[\cos \left(x\right)+\frac{5}{2}\cdot \sin \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $\cos \left(\frac{3}{2}\pi \right)+\frac{5}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(\cos \left(0\right)+\frac{5}{2}\cdot \sin \left(0\right)\right)$

= $0+\frac{5}{2}\cdot \left(-1\right)-\left(1+\frac{5}{2}\cdot 0\right)$

= $0-\frac{5}{2}-\left(1+0\right)$

= $0-\frac{5}{2}-1$

= $-\frac{5}{2}-1$

= $-\frac{5}{2}-\frac{2}{2}$

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

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

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

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

= $-\frac{2}{3}\cdot 1+\frac{2}{3}\cdot 0$

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

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

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