$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 4 = 3 ⋅ 4 = 12.

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 = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

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

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 12 +6 -2 -4 = 12

= ${\left[\frac{4}{3}{x}^3-\frac{5}{2}{x}^2\right]}_{-1}^0$

$=$ $\frac{4}{3}\cdot 0^3-\frac{5}{2}\cdot 0^2-\left(\frac{4}{3}\cdot {\left(-1\right)}^3-\frac{5}{2}\cdot {\left(-1\right)}^2\right)$

= $\frac{4}{3}\cdot 0-\frac{5}{2}\cdot 0-\left(\frac{4}{3}\cdot \left(-1\right)-\frac{5}{2}\cdot 1\right)$

= $0+0-\left(-\frac{4}{3}-\frac{5}{2}\right)$

= $0-\left(-\frac{8}{6}-\frac{15}{6}\right)$

= $-1·\left(-\frac{23}{6}\right)$

= $\frac{23}{6}$

= ${\left[\frac{2}{3}{x}^6-3x^{-1}\right]}_2^5$

= ${\left[\frac{2}{3}{x}^6-\frac{3}{x}\right]}_2^5$

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

= $\frac{2}{3}\cdot 15625-3\cdot \left(\frac{1}{5}\right)-\left(\frac{2}{3}\cdot 64-3\cdot \left(\frac{1}{2}\right)\right)$

= $\frac{31250}{3}-\frac{3}{5}-\left(\frac{128}{3}-\frac{3}{2}\right)$

= $\frac{156250}{15}-\frac{9}{15}-\left(\frac{256}{6}-\frac{9}{6}\right)$

= $\frac{156241}{15}-1·\frac{247}{6}$

= $\frac{156241}{15}-\frac{247}{6}$

= $\frac{312482}{30}-\frac{1235}{30}$

= $\frac{103749}{10}$

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

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

= $-\frac{2}{3}\cdot 2^3+3\cdot 16-\left(-\frac{2}{3}\cdot 0^3+3\cdot 4\right)$

= $-\frac{2}{3}\cdot 8+48-\left(-\frac{2}{3}\cdot 0+12\right)$

= $-\frac{16}{3}+48-\left(0+12\right)$

= $-\frac{16}{3}+\frac{144}{3}-12$

= $\frac{128}{3}-12$

= $\frac{128}{3}-\frac{36}{3}$

= $\frac{92}{3}$

= ${\left[4e^{-x}+\frac{7}{2}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $4e^{-\pi }+\frac{7}{2}\cdot \cos \left(\pi \right)-\left(4e^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{2}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $4e^{-\pi }+\frac{7}{2}\cdot \left(-1\right)-\left(4e^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{2}\cdot 0\right)$

= $4e^{-\pi }-\frac{7}{2}-\left(4e^{-\left(\frac{1}{2}\pi \right)}+0\right)$

= $4e^{-\pi }-\frac{7}{2}-4e^{-\frac{1}{2}\pi }$

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

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

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

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

= $0+0$

= 0