$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 12 +4 -1 -2 = 13

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

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

= $-\frac{3}{2}\cdot 4+10-\left(-\frac{3}{2}\cdot 1-5\right)$

= $-6+10-\left(-\frac{3}{2}-5\right)$

= $4-\left(-\frac{3}{2}-\frac{10}{2}\right)$

= $4-1·\left(-\frac{13}{2}\right)$

= $4+\frac{13}{2}$

= $\frac{8}{2}+\frac{13}{2}$

= $\frac{21}{2}$

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

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

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

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

= $\frac{4}{5}\cdot {\pi }^5-\left(-2+\frac{1}{40}\cdot {\pi }^5\right)$

= $-1·\left(-2\right)-1·\frac{1}{40}\cdot {\pi }^5+\frac{4}{5}\cdot {\pi }^5$

= $2-\frac{1}{40}\cdot {\pi }^5+\frac{4}{5}\cdot {\pi }^5$

= $2+\frac{31}{40}\cdot {\pi }^5$

= ${\left[-\frac{2}{3}{e}^{-3x+4}\right]}_0^1$

$=$ $-\frac{2}{3}{e}^{-3\cdot 1+4}+\frac{2}{3}{e}^{-3\cdot 0+4}$

= $-\frac{2}{3}{e}^{-3+4}+\frac{2}{3}{e}^{0+4}$

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

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

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

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

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

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

= $-2+\frac{5}{4}\cdot {\pi }^4-\frac{5}{64}\cdot {\pi }^4$

= $-2+\frac{75}{64}\cdot {\pi }^4$

= ${\left[\frac{2}{9}{\left(-3x+6\right)}^{\frac{3}{2}}\right]}_{-\frac{19}{3}}^{-1}$

= ${\left[\frac{2}{9}{\left(\sqrt{-3x+6}\right)}^3\right]}_{-\frac{19}{3}}^{-1}$

$=$ $\frac{2}{9}{\left(\sqrt{-3\cdot \left(-1\right)+6}\right)}^3-\frac{2}{9}{\left(\sqrt{-3\cdot \left(-\frac{19}{3}\right)+6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{3+6}\right)}^3-\frac{2}{9}{\left(\sqrt{19+6}\right)}^3$

= $\frac{2}{9}{\left(\sqrt{9}\right)}^3-\frac{2}{9}{\left(\sqrt{25}\right)}^3$

= $\frac{2}{9}\cdot 3^3-\frac{2}{9}\cdot 5^3$

= $\frac{2}{9}\cdot 27-\frac{2}{9}\cdot 125$

= $6-\frac{250}{9}$

= $\frac{54}{9}-\frac{250}{9}$

= $-\frac{196}{9}$