$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 3 = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 +3 +6 = 6

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

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

= $\frac{1}{2}\cdot 4-8-\left(\frac{1}{2}\cdot 1+4\right)$

= $2-8-\left(\frac{1}{2}+4\right)$

= $-6-\left(\frac{1}{2}+\frac{8}{2}\right)$

= $-6-1·\frac{9}{2}$

= $-6-\frac{9}{2}$

= $-\frac{12}{2}-\frac{9}{2}$

= $-\frac{21}{2}$

= ${\left[-\frac{5}{4}{e}^{2x}+4\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{4}{e}^{2\cdot \pi }+4\cdot \cos \left(\pi \right)-\left(-\frac{5}{4}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}+4\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{4}{e}^{2\cdot \pi }+4\cdot \left(-1\right)-\left(-\frac{5}{4}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}+4\cdot 0\right)$

= $-\frac{5}{4}{e}^{2\cdot \pi }-4-\left(-\frac{5}{4}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}+0\right)$

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

= $-\frac{5}{4}{e}^{2\pi }+\frac{5}{4}{e}^{\pi }-4$

= ${\left[-\frac{4}{3}{\left(-x+1\right)}^{\frac{3}{2}}\right]}_{-24}^{-8}$

= ${\left[-\frac{4}{3}{\left(\sqrt{-x+1}\right)}^3\right]}_{-24}^{-8}$

$=$ $-\frac{4}{3}{\left(\sqrt{-\left(-8\right)+1}\right)}^3+\frac{4}{3}{\left(\sqrt{-\left(-24\right)+1}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{8+1}\right)}^3+\frac{4}{3}{\left(\sqrt{24+1}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{9}\right)}^3+\frac{4}{3}{\left(\sqrt{25}\right)}^3$

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

= $-\frac{4}{3}\cdot 27+\frac{4}{3}\cdot 125$

= $-36+\frac{500}{3}$

= $-\frac{108}{3}+\frac{500}{3}$

= $\frac{392}{3}$

= ${\left[-x^{-2}+\frac{5}{4}{x}^4\right]}_2^3$

= ${\left[-\frac{1}{x^2}+\frac{5}{4}{x}^4\right]}_2^3$

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

= $-\left(\frac{1}{9}\right)+\frac{5}{4}\cdot 81-\left(-\left(\frac{1}{4}\right)+\frac{5}{4}\cdot 16\right)$

= $-\frac{1}{9}+\frac{405}{4}-\left(-\frac{1}{4}+20\right)$

= $-\frac{4}{36}+\frac{3645}{36}-\left(-\frac{1}{4}+\frac{80}{4}\right)$

= $\frac{3641}{36}-1·\frac{79}{4}$

= $\frac{3641}{36}-\frac{79}{4}$

= $\frac{3641}{36}-\frac{711}{36}$

= $\frac{1465}{18}$

= ${\left[3e^{x-2}\right]}_2^4$

$=$ $3e^{4-2}-3e^{2-2}$

= $3e^2-3e^0$

= $3e^2-3$