$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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:

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

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4 -6 -8 = -10

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

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

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

= $-6-8-\left(-6+8\right)$

= $-14-1·2$

= $-14-2$

= $-16$

= ${\left[-\frac{5}{4}{e}^{-x}+\frac{7}{4}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{4}{e}^{-\pi }+\frac{7}{4}\cdot \sin \left(\pi \right)-\left(-\frac{5}{4}{e}^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{4}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{4}{e}^{-\pi }+\frac{7}{4}\cdot 0-\left(-\frac{5}{4}{e}^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{4}\cdot 1\right)$

= $-\frac{5}{4}{e}^{-\pi }+0-\left(-\frac{5}{4}{e}^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{4}\right)$

= $-\frac{5}{4}{e}^{-\pi }-\left(-\frac{5}{4}{e}^{-\left(\frac{1}{2}\pi \right)}+\frac{7}{4}\right)$

= $-\frac{5}{4}{e}^{-\pi }+\frac{5}{4}{e}^{-\frac{1}{2}\pi }-1·\frac{7}{4}$

= $-\frac{5}{4}{e}^{-\pi }+\frac{5}{4}{e}^{-\frac{1}{2}\pi }-\frac{7}{4}$

= $\frac{5}{4}{e}^{-\frac{1}{2}\pi }-\frac{5}{4}{e}^{-\pi }-\frac{7}{4}$

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

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

= $-e^{-6+2}+e^{-2+2}$

= $-e^{-4}+e^0$

= $-e^{-4}+1$

= ${\left[-\frac{8}{3}\cdot \cos \left(x\right)+\frac{1}{3}{e}^{3x}\right]}_0^{\frac{1}{2}\pi }$

$=$ $-\frac{8}{3}\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{1}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(-\frac{8}{3}\cdot \cos \left(0\right)+\frac{1}{3}{e}^{3\cdot \left(0\right)}\right)$

= $-\frac{8}{3}\cdot 0+\frac{1}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(-\frac{8}{3}\cdot 1+\frac{1}{3}{e}^0\right)$

= $0+\frac{1}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}-\left(-\frac{8}{3}+\frac{1}{3}\right)$

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

= $\frac{1}{3}{e}^{\frac{3}{2}\pi }+\frac{7}{3}$

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

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

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

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

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

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

= $\frac{1}{3}$