$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 -4.5 +1.5 = -9

= ${\left[\frac{2}{3}{x}^3+4x\right]}_1^5$

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

= $\frac{2}{3}\cdot 125+20-\left(\frac{2}{3}\cdot 1+4\right)$

= $\frac{250}{3}+20-\left(\frac{2}{3}+4\right)$

= $\frac{250}{3}+\frac{60}{3}-\left(\frac{2}{3}+\frac{12}{3}\right)$

= $\frac{310}{3}-1·\frac{14}{3}$

= $\frac{310}{3}-\frac{14}{3}$

= $\frac{296}{3}$

= ${\left[\frac{5}{24}{x}^6+\frac{3}{2}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{5}{24}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{3}{2}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6+\frac{3}{2}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{5}{24}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{3}{2}\cdot 0-\left(\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6+\frac{3}{2}\cdot 0\right)$

= $\frac{5}{24}\cdot {\left(\frac{3}{2}\pi \right)}^6+0-\left(\frac{5}{24}\cdot {\left(\frac{1}{2}\pi \right)}^6+0\right)$

= $\frac{1215}{512}\cdot {\pi }^6-\frac{5}{1536}\cdot {\pi }^6$

= $\frac{3645}{1536}\cdot {\pi }^6-\frac{5}{1536}\cdot {\pi }^6$

= $\frac{455}{192}\cdot {\pi }^6$

= ${\left[-\frac{1}{6}{\left(-3x+6\right)}^4+6x\right]}_1^4$

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

= $-\frac{1}{6}\cdot {\left(-12+6\right)}^4+24-\left(-\frac{1}{6}\cdot {\left(-3+6\right)}^4+6\right)$

= $-\frac{1}{6}\cdot {\left(-6\right)}^4+24-\left(-\frac{1}{6}\cdot 3^4+6\right)$

= $-\frac{1}{6}\cdot 1296+24-\left(-\frac{1}{6}\cdot 81+6\right)$

= $-216+24-\left(-\frac{27}{2}+6\right)$

= $-192-\left(-\frac{27}{2}+\frac{12}{2}\right)$

= $-192-1·\left(-\frac{15}{2}\right)$

= $-192+\frac{15}{2}$

= $-\frac{384}{2}+\frac{15}{2}$

= $-\frac{369}{2}$

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

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

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

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

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

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

= $\frac{1}{3}{e}^{\frac{3}{2}\pi }-\frac{7}{3}$

= ${\left[-\frac{4}{9}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_3^7$

= ${\left[-\frac{4}{9}{\left(\sqrt{3x-5}\right)}^3\right]}_3^7$

$=$ $-\frac{4}{9}{\left(\sqrt{3\cdot 7-5}\right)}^3+\frac{4}{9}{\left(\sqrt{3\cdot 3-5}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{21-5}\right)}^3+\frac{4}{9}{\left(\sqrt{9-5}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{16}\right)}^3+\frac{4}{9}{\left(\sqrt{4}\right)}^3$

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

= $-\frac{4}{9}\cdot 64+\frac{4}{9}\cdot 8$

= $-\frac{256}{9}+\frac{32}{9}$

= $-\frac{224}{9}$