$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -12 -6 +1 = -17

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

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

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

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

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

= $\frac{5}{2}-1·\frac{1}{2}$

= $\frac{5}{2}-\frac{1}{2}$

= $2$

= ${\left[\frac{27}{16}{x}^{\frac{4}{3}}-\frac{3}{4}{x}^3\right]}_0^1$

= ${\left[\frac{27}{16}{\left(\sqrt[3]{x}\right)}^4-\frac{3}{4}{x}^3\right]}_0^1$

$=$ $\frac{27}{16}{\left(\sqrt[3]{1}\right)}^4-\frac{3}{4}\cdot 1^3-\left(\frac{27}{16}{\left(\sqrt[3]{0}\right)}^4-\frac{3}{4}\cdot 0^3\right)$

= $\frac{27}{16}\cdot 1^4-\frac{3}{4}\cdot 1-\left(\frac{27}{16}\cdot 0^4-\frac{3}{4}\cdot 0\right)$

= $\frac{27}{16}\cdot 1-\frac{3}{4}-\left(\frac{27}{16}\cdot 0+0\right)$

= $\frac{27}{16}-\frac{3}{4}-\left(0+0\right)$

= $\frac{27}{16}-\frac{12}{16}+0$

= $\frac{15}{16}$

= ${\left[\frac{1}{8}{\left(2x-3\right)}^4-x\right]}_2^4$

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

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

= $\frac{1}{8}\cdot 5^4-4-\left(\frac{1}{8}\cdot 1^4-2\right)$

= $\frac{1}{8}\cdot 625-4-\left(\frac{1}{8}\cdot 1-2\right)$

= $\frac{625}{8}-4-\left(\frac{1}{8}-2\right)$

= $\frac{625}{8}-\frac{32}{8}-\left(\frac{1}{8}-\frac{16}{8}\right)$

= $\frac{593}{8}-1·\left(-\frac{15}{8}\right)$

= $\frac{593}{8}+\frac{15}{8}$

= $76$

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

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

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

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

= $3+3$

= $6$

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

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

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

= $\frac{1}{2}\cdot 1-\frac{1}{2}\cdot 2^4$

= $\frac{1}{2}-\frac{1}{2}\cdot 16$

= $\frac{1}{2}-8$

= $\frac{1}{2}-\frac{16}{2}$

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