$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : 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:

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

= ${\left[\frac{5}{2}{x}^2-2x\right]}_1^4$

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

= $\frac{5}{2}\cdot 16-8-\left(\frac{5}{2}\cdot 1-2\right)$

= $40-8-\left(\frac{5}{2}-2\right)$

= $32-\left(\frac{5}{2}-\frac{4}{2}\right)$

= $32-1·\frac{1}{2}$

= $32-\frac{1}{2}$

= $\frac{64}{2}-\frac{1}{2}$

= $\frac{63}{2}$

= ${\left[3e^{-x}+\frac{2}{3}{x}^{\frac{3}{2}}\right]}_4^{16}$

= ${\left[3e^{-x}+\frac{2}{3}{\left(\sqrt{x}\right)}^3\right]}_4^{16}$

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

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

= $3e^{-16}+\frac{2}{3}\cdot 64-\left(3e^{-4}+\frac{2}{3}\cdot 8\right)$

= $3e^{-16}+\frac{128}{3}-\left(3e^{-4}+\frac{16}{3}\right)$

= $-3e^{-4}-1·\frac{16}{3}+3e^{-16}+\frac{128}{3}$

= $-3e^{-4}-\frac{16}{3}+3e^{-16}+\frac{128}{3}$

= $-3e^{-4}+3e^{-16}-\frac{16}{3}+\frac{128}{3}$

= $-3e^{-4}+3e^{-16}+\frac{112}{3}$

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

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

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

= $\frac{2}{3}\cdot 1-\frac{2}{3}\cdot 0$

= $\frac{2}{3}+0$

= $\frac{2}{3}+0$

= $\frac{2}{3}$

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

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

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

= $4e^{\pi }-\frac{3}{4}-\left(4e^{\frac{1}{2}\pi }+0\right)$

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

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

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

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

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

= $\frac{1}{2}\cdot 0-\frac{1}{2}\cdot 0$

= $0+0$

= 0