$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 6) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 +1.5 +2 = -2.5

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

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

= $\frac{5}{2}\cdot 9-6-\left(\frac{5}{2}\cdot 1-2\right)$

= $\frac{45}{2}-6-\left(\frac{5}{2}-2\right)$

= $\frac{45}{2}-\frac{12}{2}-\left(\frac{5}{2}-\frac{4}{2}\right)$

= $\frac{33}{2}-1·\frac{1}{2}$

= $\frac{33}{2}-\frac{1}{2}$

= $16$

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

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

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

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

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

= $-\frac{3}{2}+\frac{1}{{\pi }^2}-\frac{4}{{\pi }^2}$

= $-\frac{3}{2}-\frac{3}{{\pi }^2}$

= ${\left[\frac{1}{4}{\left(3x-7\right)}^4-2x\right]}_2^3$

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

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

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

= $\frac{1}{4}\cdot 16-6-\left(\frac{1}{4}\cdot 1-4\right)$

= $4-6-\left(\frac{1}{4}-4\right)$

= $-2-\left(\frac{1}{4}-\frac{16}{4}\right)$

= $-2-1·\left(-\frac{15}{4}\right)$

= $-2+\frac{15}{4}$

= $-\frac{8}{4}+\frac{15}{4}$

= $\frac{7}{4}$

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

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

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

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

= $\frac{1}{3{\pi }^3}-1·\left(-1\right)-1·\frac{8}{3{\pi }^3}$

= $\frac{1}{3{\pi }^3}+1-\frac{8}{3{\pi }^3}$

= $1+\frac{1}{3{\pi }^3}-\frac{8}{3{\pi }^3}$

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

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

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

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

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

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

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

= $\frac{16}{3}$