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

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

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 -2 -4 = -3

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

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

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

= $-6+2-\left(-6-2\right)$

= $-4-1·\left(-8\right)$

= $-4+8$

= $4$

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

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

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

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

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

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

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

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

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

= ${\left[e^{2x-1}\right]}_2^4$

$=$ $e^{2\cdot 4-1}-e^{2\cdot 2-1}$

= $e^{8-1}-e^{4-1}$

= $e^7-e^3$

= ${\left[-5\cdot \cos \left(x\right)+x^5\right]}_0^{\pi }$

$=$ $-5\cdot \cos \left(\pi \right)+{\pi }^5-\left(-5\cdot \cos \left(0\right)+{\left(0\right)}^5\right)$

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

= $5+{\pi }^5-\left(-5+0\right)$

= $5+{\pi }^5+5$

= $5+5+{\pi }^5$

= $10+{\pi }^5$

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

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

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

= $-2\cdot 1+2\cdot \left(-1\right)$

= $-2-2$

= $-4$