$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ 3 = 2 ⋅ 3 = 6.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 6 +3 -4.5 -6 = -1.5

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

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

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

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

= $\frac{4}{6}+\frac{3}{6}-\left(-\frac{4}{6}+\frac{3}{6}\right)$

= $\frac{7}{6}-1·\left(-\frac{1}{6}\right)$

= $\frac{7}{6}+\frac{1}{6}$

= $\frac{4}{3}$

= ${\left[-\frac{3}{5}{x}^5+\frac{1}{2}\cdot \sin \left(x\right)\right]}_0^{\pi }$

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

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

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

= $-\frac{3}{5}\cdot {\pi }^5+0$

= $-\frac{3}{5}\cdot {\pi }^5$

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

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

= $\cos \left(\frac{7}{2}\pi \right)-\cos (-\pi )$

= $0-\left(-1\right)$

= $0+1$

= $1$

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

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

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

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

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

= $\frac{1}{2}{e}^{-\frac{9}{2}\pi }-2-\frac{1}{2}{e}^{-\frac{3}{2}\pi }-1·2$

= $\frac{1}{2}{e}^{-\frac{9}{2}\pi }-2-\frac{1}{2}{e}^{-\frac{3}{2}\pi }-2$

= $-\frac{1}{2}{e}^{-\frac{3}{2}\pi }+\frac{1}{2}{e}^{-\frac{9}{2}\pi }-2-2$

= $-\frac{1}{2}{e}^{-\frac{3}{2}\pi }+\frac{1}{2}{e}^{-\frac{9}{2}\pi }-4$

= ${\left[-{\left(-x+2\right)}^3\right]}_0^3$

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

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

= $-\left(-1\right)+2^3$

= $1+8$

= $9$