$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4.5 +2 = -2.5

= ${\left[\frac{4}{3}{x}^3-\frac{5}{2}{x}^2\right]}_{-3}^0$

$=$ $\frac{4}{3}\cdot 0^3-\frac{5}{2}\cdot 0^2-\left(\frac{4}{3}\cdot {\left(-3\right)}^3-\frac{5}{2}\cdot {\left(-3\right)}^2\right)$

= $\frac{4}{3}\cdot 0-\frac{5}{2}\cdot 0-\left(\frac{4}{3}\cdot \left(-27\right)-\frac{5}{2}\cdot 9\right)$

= $0+0-\left(-36-\frac{45}{2}\right)$

= $0-\left(-\frac{72}{2}-\frac{45}{2}\right)$

= $-1·\left(-\frac{117}{2}\right)$

= $\frac{117}{2}$

= ${\left[-x^{\frac{5}{2}}+\frac{2}{9}{x}^{-3}\right]}_1^9$

= ${\left[-{\left(\sqrt{x}\right)}^5+\frac{2}{9x^3}\right]}_1^9$

$=$ $-{\left(\sqrt{9}\right)}^5+\frac{2}{9\cdot 9^3}-\left(-{\left(\sqrt{1}\right)}^5+\frac{2}{9\cdot 1^3}\right)$

= $-3^5+\frac{2}{9}\cdot \left(\frac{1}{729}\right)-\left(-1^5+\frac{2}{9}\cdot 1\right)$

= $-243+\frac{2}{6561}-\left(-1+\frac{2}{9}\right)$

= $-243+\frac{2}{6561}-1·\left(-1\right)-1·\frac{2}{9}$

= $-243+\frac{2}{6561}+1-\frac{2}{9}$

= $-\frac{1594323}{6561}+\frac{2}{6561}+\frac{6561}{6561}-\frac{1458}{6561}$

= $-\frac{1589218}{6561}$

= ${\left[-{\left(-2x+2\right)}^{\frac{3}{2}}\right]}_{-\frac{23}{2}}^{-\frac{7}{2}}$

= ${\left[-{\left(\sqrt{-2x+2}\right)}^3\right]}_{-\frac{23}{2}}^{-\frac{7}{2}}$

$=$ $-{\left(\sqrt{-2\cdot \left(-\frac{7}{2}\right)+2}\right)}^3+{\left(\sqrt{-2\cdot \left(-\frac{23}{2}\right)+2}\right)}^3$

= $-{\left(\sqrt{7+2}\right)}^3+{\left(\sqrt{23+2}\right)}^3$

= $-{\left(\sqrt{9}\right)}^3+{\left(\sqrt{25}\right)}^3$

= $-3^3+5^3$

= $-27+125$

= $98$

= ${\left[-3\cdot \cos \left(x\right)-x^3\right]}_0^{\pi }$

$=$ $-3\cdot \cos \left(\pi \right)-{\pi }^3-\left(-3\cdot \cos \left(0\right)-{\left(0\right)}^3\right)$

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

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

= $3-{\pi }^3+3$

= $3+3-{\pi }^3$

= $6-{\pi }^3$

= ${\left[\frac{4}{9}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_{\frac{7}{3}}^{\frac{28}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-3}\right)}^3\right]}_{\frac{7}{3}}^{\frac{28}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{28}{3}\right)-3}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{7}{3}\right)-3}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{28-3}\right)}^3-\frac{4}{9}{\left(\sqrt{7-3}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{25}\right)}^3-\frac{4}{9}{\left(\sqrt{4}\right)}^3$

= $\frac{4}{9}\cdot 5^3-\frac{4}{9}\cdot 2^3$

= $\frac{4}{9}\cdot 125-\frac{4}{9}\cdot 8$

= $\frac{500}{9}-\frac{32}{9}$

= $52$