$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 1 = 2 ⋅ 1 = 2.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{1\; +\; <mn>2</mn>}}{2}$ = 2 ⋅ 1.5 = 3.

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 +1 +2 +3 = 3

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

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

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

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

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

= $\frac{40}{3}-1·\frac{14}{3}$

= $\frac{40}{3}-\frac{14}{3}$

= $\frac{26}{3}$

= ${\left[-5\cdot \sin \left(x\right)+\frac{1}{5}{x}^5\right]}_0^{\frac{3}{2}\pi }$

$=$ $-5\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{1}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(-5\cdot \sin \left(0\right)+\frac{1}{5}\cdot {\left(0\right)}^5\right)$

= $-5\cdot \left(-1\right)+\frac{1}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(-5\cdot 0+\frac{1}{5}\cdot 0\right)$

= $5+\frac{1}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5-\left(0+0\right)$

= $5+\frac{243}{160}\cdot {\pi }^5+0$

= $5+\frac{243}{160}\cdot {\pi }^5$

= ${\left[-\frac{3}{2}{\left(2x-2\right)}^{-1}\right]}_3^5$

= ${\left[-\frac{3}{2\left(2x-2\right)}\right]}_3^5$

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

= $-\frac{3}{2\left(10-2\right)}+\frac{3}{2\left(6-2\right)}$

= $-\frac{3}{2\cdot 8}+\frac{3}{2\cdot 4}$

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

= $-\frac{3}{16}+\frac{3}{8}$

= $-\frac{3}{16}+\frac{6}{16}$

= $\frac{3}{16}$

= ${\left[-e^{-x}+\frac{3}{10}{x}^5\right]}_{-1}^1$

$=$ $-e^{-1}+\frac{3}{10}\cdot 1^5-\left(-e^{-\left(-1\right)}+\frac{3}{10}\cdot {\left(-1\right)}^5\right)$

= $-e^{-1}+\frac{3}{10}\cdot 1-(-e+\frac{3}{10}\cdot \left(-1\right))$

= $-e^{-1}+\frac{3}{10}-(-e-\frac{3}{10})$

= $-e^{-1}+\frac{3}{10}-(-\frac{3}{10}-e)$

= $-e^{-1}+\frac{3}{10}-1·\left(-\frac{3}{10}\right)-1·(-e)$

= $-e^{-1}+\frac{3}{10}+\frac{3}{10}+e$

= $-e^{-1}+\frac{3}{5}+e$

= ${\left[-\frac{1}{3}{\left(-2x+5\right)}^{\frac{3}{2}}\right]}_{-10}^{-\frac{11}{2}}$

= ${\left[-\frac{1}{3}{\left(\sqrt{-2x+5}\right)}^3\right]}_{-10}^{-\frac{11}{2}}$

$=$ $-\frac{1}{3}{\left(\sqrt{-2\cdot \left(-\frac{11}{2}\right)+5}\right)}^3+\frac{1}{3}{\left(\sqrt{-2\cdot \left(-10\right)+5}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{11+5}\right)}^3+\frac{1}{3}{\left(\sqrt{20+5}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{16}\right)}^3+\frac{1}{3}{\left(\sqrt{25}\right)}^3$

= $-\frac{1}{3}\cdot 4^3+\frac{1}{3}\cdot 5^3$

= $-\frac{1}{3}\cdot 64+\frac{1}{3}\cdot 125$

= $-\frac{64}{3}+\frac{125}{3}$

= $\frac{61}{3}$