$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ gibt des orientierten (also mit Vorzeichen behafteten) Flächeninhalt zwischen dem Graph der Funktion und der x-Achse an.

Um diesen aus dem Schaubild zu berechnen unterteilen wir die Fläche in Rechtecke, Parallelogrammen und ggf. Dreiecke:

I₁ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 3 = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 -3 +4.5 +6 = 3.5

= ${\left[-\frac{1}{2}{x}^2-3x\right]}_0^4$

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

= $-\frac{1}{2}\cdot 16-12-\left(-\frac{1}{2}\cdot 0+0\right)$

= $-8-12-\left(0+0\right)$

= $-20+0$

= $-20$

= ${\left[9\cdot \sin \left(x\right)-\frac{2}{3}{x}^6\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $9\cdot \sin \left(\pi \right)-\frac{2}{3}\cdot {\pi }^6-\left(9\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{2}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6\right)$

= $9\cdot 0-\frac{2}{3}\cdot {\pi }^6-\left(9\cdot 1-\frac{2}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6\right)$

= $0-\frac{2}{3}\cdot {\pi }^6-\left(9-\frac{2}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6\right)$

= $-\frac{2}{3}\cdot {\pi }^6-\left(9-\frac{1}{96}\cdot {\pi }^6\right)$

= $-9+\frac{1}{96}\cdot {\pi }^6-\frac{2}{3}\cdot {\pi }^6$

= $-9-\frac{21}{32}\cdot {\pi }^6$

= ${\left[-2e^{x-2}\right]}_2^5$

$=$ $-2e^{5-2}+2e^{2-2}$

= $-2e^3+2e^0$

= $-2e^3+2$

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

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

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

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

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

= $\frac{1}{24{\pi }^3}-\left(3+\frac{8}{3{\pi }^3}\right)$

= $-3-\frac{8}{3{\pi }^3}+\frac{1}{24{\pi }^3}$

= $-3-\frac{21}{8{\pi }^3}$

= ${\left[e^{-2x+4}\right]}_2^3$

$=$ $e^{-2\cdot 3+4}-e^{-2\cdot 2+4}$

= $e^{-6+4}-e^{-4+4}$

= $e^{-2}-e^0$

= $e^{-2}-1$