$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

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

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-2\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 3 -2 -6 -5 = -10

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

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

= $-\frac{2}{3}\cdot \left(-1\right)+5-\left(-\frac{2}{3}\cdot \left(-8\right)+10\right)$

= $\frac{2}{3}+5-\left(\frac{16}{3}+10\right)$

= $\frac{2}{3}+\frac{15}{3}-\left(\frac{16}{3}+\frac{30}{3}\right)$

= $\frac{17}{3}-1·\frac{46}{3}$

= $\frac{17}{3}-\frac{46}{3}$

= $-\frac{29}{3}$

= ${\left[\frac{9}{4}\cdot \cos \left(x\right)+\sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{9}{4}\cdot \cos \left(\pi \right)+\sin \left(\pi \right)-\left(\frac{9}{4}\cdot \cos \left(\frac{1}{2}\pi \right)+\sin \left(\frac{1}{2}\pi \right)\right)$

= $\frac{9}{4}\cdot \left(-1\right)+0-\left(\frac{9}{4}\cdot 0+1\right)$

= $-\frac{9}{4}+0-\left(0+1\right)$

= $-\frac{9}{4}+0-1$

= $-\frac{9}{4}-1$

= $-\frac{9}{4}-\frac{4}{4}$

= $-\frac{13}{4}$

= ${\left[\frac{1}{4}{\left(-2x+1\right)}^{-2}\right]}_2^4$

= ${\left[\frac{1}{4{\left(-2x+1\right)}^2}\right]}_2^4$

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

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

= $\frac{1}{4{\left(-7\right)}^2}-\frac{1}{4{\left(-3\right)}^2}$

= $\frac{1}{4}\cdot \left(\frac{1}{49}\right)-\frac{1}{4}\cdot \left(\frac{1}{9}\right)$

= $\frac{1}{196}-\frac{1}{36}$

= $\frac{9}{1764}-\frac{49}{1764}$

= $-\frac{10}{441}$

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

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

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

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

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

= ${\left[\frac{1}{3}{e}^{-3x+6}\right]}_0^2$

$=$ $\frac{1}{3}{e}^{-3\cdot 2+6}-\frac{1}{3}{e}^{-3\cdot 0+6}$

= $\frac{1}{3}{e}^{-6+6}-\frac{1}{3}{e}^{0+6}$

= $\frac{1}{3}{e}^0-\frac{1}{3}{e}^6$

= $\frac{1}{3}-\frac{1}{3}{e}^6$