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

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

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 -2 = 1

= ${\left[-\frac{4}{3}{x}^3-5x\right]}_1^3$

$=$ $-\frac{4}{3}\cdot 3^3-5\cdot 3-\left(-\frac{4}{3}\cdot 1^3-5\cdot 1\right)$

= $-\frac{4}{3}\cdot 27-15-\left(-\frac{4}{3}\cdot 1-5\right)$

= $-36-15-\left(-\frac{4}{3}-5\right)$

= $-51-\left(-\frac{4}{3}-\frac{15}{3}\right)$

= $-51-1·\left(-\frac{19}{3}\right)$

= $-51+\frac{19}{3}$

= $-\frac{153}{3}+\frac{19}{3}$

= $-\frac{134}{3}$

= ${\left[5e^x+\frac{5}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

= $5e^{\pi }-\frac{5}{3}-5e^{\frac{1}{2}\pi }$

= $5e^{\pi }-5e^{\frac{1}{2}\pi }-\frac{5}{3}$

= ${\left[-\frac{1}{6}{\left(3x-7\right)}^4-x^2\right]}_0^3$

$=$ $-\frac{1}{6}\cdot {\left(3\cdot 3-7\right)}^4-3^2-\left(-\frac{1}{6}\cdot {\left(3\cdot 0-7\right)}^4-0^2\right)$

= $-\frac{1}{6}\cdot {\left(9-7\right)}^4-9-\left(-\frac{1}{6}\cdot {\left(0-7\right)}^4-0\right)$

= $-\frac{1}{6}\cdot 2^4-9-\left(-\frac{1}{6}\cdot {\left(-7\right)}^4+0\right)$

= $-\frac{1}{6}\cdot 16-9-\left(-\frac{1}{6}\cdot 2401+0\right)$

= $-\frac{8}{3}-9-\left(-\frac{2401}{6}+0\right)$

= $-\frac{8}{3}-\frac{27}{3}-\left(-\frac{2401}{6}+0\right)$

= $-\frac{35}{3}+\frac{2401}{6}$

= $-\frac{70}{6}+\frac{2401}{6}$

= $-\frac{35}{3}+\frac{2401}{6}$

= $-\frac{70}{6}+\frac{2401}{6}$

= $\frac{777}{2}$

= ${\left[-4\cdot \cos \left(x\right)+3\cdot \sin \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $-4\cdot \cos \left(\frac{3}{2}\pi \right)+3\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-4\cdot \cos \left(0\right)+3\cdot \sin \left(0\right)\right)$

= $-4\cdot 0+3\cdot \left(-1\right)-\left(-4\cdot 1+3\cdot 0\right)$

= $0-3-\left(-4+0\right)$

= $-3+4$

= $1$

= ${\left[-\frac{2}{3}{\left(-3x+5\right)}^{\frac{3}{2}}\right]}_{-\frac{4}{3}}^{\frac{1}{3}}$

= ${\left[-\frac{2}{3}{\left(\sqrt{-3x+5}\right)}^3\right]}_{-\frac{4}{3}}^{\frac{1}{3}}$

$=$ $-\frac{2}{3}{\left(\sqrt{-3\cdot \left(\frac{1}{3}\right)+5}\right)}^3+\frac{2}{3}{\left(\sqrt{-3\cdot \left(-\frac{4}{3}\right)+5}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{-1+5}\right)}^3+\frac{2}{3}{\left(\sqrt{4+5}\right)}^3$

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

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

= $-\frac{2}{3}\cdot 8+\frac{2}{3}\cdot 27$

= $-\frac{16}{3}+18$

= $-\frac{16}{3}+\frac{54}{3}$

= $\frac{38}{3}$