$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ 2 = 3 ⋅ 2 = 6.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

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

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 6 +2 -6 -8 = -6

= ${\left[\frac{2}{3}{x}^3-\frac{5}{2}{x}^2\right]}_0^1$

$=$ $\frac{2}{3}\cdot 1^3-\frac{5}{2}\cdot 1^2-\left(\frac{2}{3}\cdot 0^3-\frac{5}{2}\cdot 0^2\right)$

= $\frac{2}{3}\cdot 1-\frac{5}{2}\cdot 1-\left(\frac{2}{3}\cdot 0-\frac{5}{2}\cdot 0\right)$

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

= $\frac{4}{6}-\frac{15}{6}+0$

= $-\frac{11}{6}$

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

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

= $-3\cdot 0+\frac{1}{2}\cdot {\pi }^4-\left(-3\cdot 1+\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $0+\frac{1}{2}\cdot {\pi }^4-\left(-3+\frac{1}{2}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $\frac{1}{2}\cdot {\pi }^4-\left(-3+\frac{1}{32}\cdot {\pi }^4\right)$

= $-1·\left(-3\right)-1·\frac{1}{32}\cdot {\pi }^4+\frac{1}{2}\cdot {\pi }^4$

= $3-\frac{1}{32}\cdot {\pi }^4+\frac{1}{2}\cdot {\pi }^4$

= $3+\frac{15}{32}\cdot {\pi }^4$

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

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

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

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

= $\frac{1}{3}+0$

= $\frac{1}{3}+0$

= $\frac{1}{3}$

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

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

= $\frac{1}{5}\cdot 243-\frac{3}{4}{e}^{-9}-\left(\frac{1}{5}\cdot 0-\frac{3}{4}{e}^0\right)$

= $\frac{243}{5}-\frac{3}{4}{e}^{-9}-\left(0-\frac{3}{4}\right)$

= $-\frac{3}{4}{e}^{-9}+\frac{243}{5}-\left(0-\frac{3}{4}\right)$

= $-\frac{3}{4}{e}^{-9}+\frac{243}{5}+\frac{3}{4}$

= $-\frac{3}{4}{e}^{-9}+\frac{987}{20}$

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

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

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

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

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

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

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