$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

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 = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 4 = 2 ⋅ 4 = 8.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{4\; +\; <mn>1</mn>}}{2}$ = 2 ⋅ 2.5 = 5.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -3 +4 +8 +5 = 14

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

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

= $-\frac{4}{3}\cdot 1+\frac{3}{2}\cdot 1-\left(-\frac{4}{3}\cdot \left(-27\right)+\frac{3}{2}\cdot 9\right)$

= $-\frac{4}{3}+\frac{3}{2}-\left(36+\frac{27}{2}\right)$

= $-\frac{8}{6}+\frac{9}{6}-\left(\frac{72}{2}+\frac{27}{2}\right)$

= $\frac{1}{6}-1·\frac{99}{2}$

= $\frac{1}{6}-\frac{99}{2}$

= $\frac{1}{6}-\frac{297}{6}$

= $-\frac{148}{3}$

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

= ${\left[-\frac{1}{3}{\left(\sqrt{x}\right)}^3+\frac{1}{3}{x}^3\right]}_0^4$

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

= $-\frac{1}{3}\cdot 2^3+\frac{1}{3}\cdot 64-\left(-\frac{1}{3}\cdot 0^3+\frac{1}{3}\cdot 0\right)$

= $-\frac{1}{3}\cdot 8+\frac{64}{3}-\left(-\frac{1}{3}\cdot 0+0\right)$

= $-\frac{8}{3}+\frac{64}{3}-\left(0+0\right)$

= $\frac{56}{3}+0$

= $\frac{56}{3}$

= ${\left[-\frac{2}{9}{\left(3x-4\right)}^{-3}\right]}_2^4$

= ${\left[-\frac{2}{9{\left(3x-4\right)}^3}\right]}_2^4$

$=$ $-\frac{2}{9{\left(3\cdot 4-4\right)}^3}+\frac{2}{9{\left(3\cdot 2-4\right)}^3}$

= $-\frac{2}{9{\left(12-4\right)}^3}+\frac{2}{9{\left(6-4\right)}^3}$

= $-\frac{2}{9\cdot 8^3}+\frac{2}{9\cdot 2^3}$

= $-\frac{2}{9}\cdot \left(\frac{1}{512}\right)+\frac{2}{9}\cdot \left(\frac{1}{8}\right)$

= $-\frac{1}{2304}+\frac{1}{36}$

= $-\frac{1}{2304}+\frac{64}{2304}$

= $\frac{7}{256}$

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

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

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

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

= $-\frac{3}{2}+\frac{1}{3}\cdot {\pi }^3-\frac{1}{24}\cdot {\pi }^3$

= $-\frac{3}{2}+\frac{7}{24}\cdot {\pi }^3$

= ${\left[-e^{x-1}\right]}_1^3$

$=$ $-e^{3-1}+e^{1-1}$

= $-e^2+e^0$

= $-e^2+1$