$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

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

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

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 6 -4 -8 = -6

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

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

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

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

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

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

= $\frac{1}{2}+\frac{3}{2}$

= $2$

= ${\left[-\frac{9}{2}\cdot \sin \left(x\right)+\frac{5}{3}{x}^3\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{9}{2}\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{5}{3}\cdot {\left(\frac{3}{2}\pi \right)}^3-\left(-\frac{9}{2}\cdot \sin \left(\frac{1}{2}\pi \right)+\frac{5}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3\right)$

= $-\frac{9}{2}\cdot \left(-1\right)+\frac{5}{3}\cdot {\left(\frac{3}{2}\pi \right)}^3-\left(-\frac{9}{2}\cdot 1+\frac{5}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3\right)$

= $\frac{9}{2}+\frac{5}{3}\cdot {\left(\frac{3}{2}\pi \right)}^3-\left(-\frac{9}{2}+\frac{5}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3\right)$

= $\frac{9}{2}+\frac{45}{8}\cdot {\pi }^3-\left(-\frac{9}{2}+\frac{5}{24}\cdot {\pi }^3\right)$

= $\frac{9}{2}+\frac{45}{8}\cdot {\pi }^3-1·\left(-\frac{9}{2}\right)-1·\frac{5}{24}\cdot {\pi }^3$

= $\frac{9}{2}+\frac{45}{8}\cdot {\pi }^3+\frac{9}{2}-\frac{5}{24}\cdot {\pi }^3$

= $\frac{9}{2}+\frac{9}{2}+\frac{45}{8}\cdot {\pi }^3-\frac{5}{24}\cdot {\pi }^3$

= $9+\frac{65}{12}\cdot {\pi }^3$

= ${\left[-\frac{1}{6}{\left(-2x+5\right)}^3-4x\right]}_1^2$

$=$ $-\frac{1}{6}\cdot {\left(-2\cdot 2+5\right)}^3-4\cdot 2-\left(-\frac{1}{6}\cdot {\left(-2\cdot 1+5\right)}^3-4\cdot 1\right)$

= $-\frac{1}{6}\cdot {\left(-4+5\right)}^3-8-\left(-\frac{1}{6}\cdot {\left(-2+5\right)}^3-4\right)$

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

= $-\frac{1}{6}\cdot 1-8-\left(-\frac{1}{6}\cdot 27-4\right)$

= $-\frac{1}{6}-8-\left(-\frac{9}{2}-4\right)$

= $-\frac{1}{6}-\frac{48}{6}-\left(-\frac{9}{2}-\frac{8}{2}\right)$

= $-\frac{49}{6}-1·\left(-\frac{17}{2}\right)$

= $-\frac{49}{6}+\frac{17}{2}$

= $-\frac{49}{6}+\frac{51}{6}$

= $\frac{1}{3}$

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

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

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

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

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

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

= $-3e^{\frac{3}{2}\pi }+2+3e^{\frac{1}{2}\pi }+2$

= $-3e^{\frac{3}{2}\pi }+3e^{\frac{1}{2}\pi }+2+2$

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

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

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

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

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

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

= $-1$