$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

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

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 6 -4 -8 = -6

= ${\left[-\frac{2}{3}{x}^3+4x\right]}_1^2$

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

= $-\frac{2}{3}\cdot 8+8-\left(-\frac{2}{3}\cdot 1+4\right)$

= $-\frac{16}{3}+8-\left(-\frac{2}{3}+4\right)$

= $-\frac{16}{3}+\frac{24}{3}-\left(-\frac{2}{3}+\frac{12}{3}\right)$

= $\frac{8}{3}-1·\frac{10}{3}$

= $\frac{8}{3}-\frac{10}{3}$

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

= ${\left[\frac{1}{3}\cdot \sin \left(x\right)+4\cdot \cos \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

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

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

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

= $-\frac{1}{3}+0-4$

= $-\frac{1}{3}-4$

= $-\frac{1}{3}-\frac{12}{3}$

= $-\frac{13}{3}$

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

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

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

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

= $1+1$

= $2$

= ${\left[-\frac{4}{3}{e}^{3x}+\frac{8}{3}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{4}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}+\frac{8}{3}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{4}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}+\frac{8}{3}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{4}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}+\frac{8}{3}\cdot \left(-1\right)-\left(-\frac{4}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}+\frac{8}{3}\cdot 1\right)$

= $-\frac{4}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\frac{8}{3}-\left(-\frac{4}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}+\frac{8}{3}\right)$

= $-\frac{4}{3}{e}^{\frac{9}{2}\pi }-\frac{8}{3}+\frac{4}{3}{e}^{\frac{3}{2}\pi }-1·\frac{8}{3}$

= $-\frac{4}{3}{e}^{\frac{9}{2}\pi }-\frac{8}{3}+\frac{4}{3}{e}^{\frac{3}{2}\pi }-\frac{8}{3}$

= $-\frac{4}{3}{e}^{\frac{9}{2}\pi }+\frac{4}{3}{e}^{\frac{3}{2}\pi }-\frac{8}{3}-\frac{8}{3}$

= $-\frac{4}{3}{e}^{\frac{9}{2}\pi }+\frac{4}{3}{e}^{\frac{3}{2}\pi }-\frac{16}{3}$

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

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

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

= $\frac{2}{3}\cdot 27-\frac{2}{3}\cdot 0$

= $18+0$

= $18$