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

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

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 3 = 2 ⋅ 3 = 6.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{3\; +\; <mn>4</mn>}}{2}$ = 2 ⋅ 3.5 = 7.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -2 +3 +6 +7 = 14

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

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

= $-\frac{2}{3}\cdot 8+4-\left(-\frac{2}{3}\cdot 0+0\right)$

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

= $-\frac{16}{3}+\frac{12}{3}+0$

= $-\frac{4}{3}$

= ${\left[\frac{1}{2}{x}^4+\frac{8}{3}{x}^{\frac{3}{2}}\right]}_4^9$

= ${\left[\frac{1}{2}{x}^4+\frac{8}{3}{\left(\sqrt{x}\right)}^3\right]}_4^9$

$=$ $\frac{1}{2}\cdot 9^4+\frac{8}{3}{\left(\sqrt{9}\right)}^3-\left(\frac{1}{2}\cdot 4^4+\frac{8}{3}{\left(\sqrt{4}\right)}^3\right)$

= $\frac{1}{2}\cdot 6561+\frac{8}{3}\cdot 3^3-\left(\frac{1}{2}\cdot 256+\frac{8}{3}\cdot 2^3\right)$

= $\frac{6561}{2}+\frac{8}{3}\cdot 27-\left(128+\frac{8}{3}\cdot 8\right)$

= $\frac{6561}{2}+72-\left(128+\frac{64}{3}\right)$

= $\frac{6561}{2}+\frac{144}{2}-\left(\frac{384}{3}+\frac{64}{3}\right)$

= $\frac{6705}{2}-1·\frac{448}{3}$

= $\frac{6705}{2}-\frac{448}{3}$

= $\frac{20115}{6}-\frac{896}{6}$

= $\frac{6705}{2}-\frac{448}{3}$

= $\frac{20115}{6}-\frac{896}{6}$

= $\frac{19219}{6}$

= ${\left[3e^{x-2}\right]}_2^3$

$=$ $3e^{3-2}-3e^{2-2}$

= $3e-3e^0$

= $3e-3$

= ${\left[\frac{7}{6}{x}^3-2\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $2+\frac{63}{16}\cdot {\pi }^3-\left(-2+\frac{7}{48}\cdot {\pi }^3\right)$

= $2+\frac{63}{16}\cdot {\pi }^3-1·\left(-2\right)-1·\frac{7}{48}\cdot {\pi }^3$

= $2+\frac{63}{16}\cdot {\pi }^3+2-\frac{7}{48}\cdot {\pi }^3$

= $2+2+\frac{63}{16}\cdot {\pi }^3-\frac{7}{48}\cdot {\pi }^3$

= $4+\frac{91}{24}\cdot {\pi }^3$

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

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

= $-\frac{3}{2}{e}^{-8+2}+\frac{3}{2}{e}^{-2+2}$

= $-\frac{3}{2}{e}^{-6}+\frac{3}{2}{e}^0$

= $-\frac{3}{2}{e}^{-6}+\frac{3}{2}$