$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : 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:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 +6 +7 = 16

= ${\left[-\frac{5}{3}{x}^3+x\right]}_0^3$

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

= $-\frac{5}{3}\cdot 27+3-\left(-\frac{5}{3}\cdot 0+0\right)$

= $-45+3-\left(0+0\right)$

= $-42+0$

= $-42$

= ${\left[x^3-\frac{1}{2}{e}^{2x}\right]}_{-3}^0$

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

= $0-\frac{1}{2}{e}^0-\left(\left(-27\right)-\frac{1}{2}{e}^{-6}\right)$

= $0-\frac{1}{2}-\left(-27-\frac{1}{2}{e}^{-6}\right)$

= $0-\frac{1}{2}-\left(-\frac{1}{2}{e}^{-6}-27\right)$

= $-\frac{1}{2}-\left(-\frac{1}{2}{e}^{-6}-27\right)$

= $\frac{1}{2}{e}^{-6}-1·\left(-27\right)-\frac{1}{2}$

= $\frac{1}{2}{e}^{-6}+27-\frac{1}{2}$

= $\frac{1}{2}{e}^{-6}+\frac{53}{2}$

= ${\left[\frac{1}{9}{\left(3x-5\right)}^3\right]}_0^1$

$=$ $\frac{1}{9}\cdot {\left(3\cdot 1-5\right)}^3-\frac{1}{9}\cdot {\left(3\cdot 0-5\right)}^3$

= $\frac{1}{9}\cdot {\left(3-5\right)}^3-\frac{1}{9}\cdot {\left(0-5\right)}^3$

= $\frac{1}{9}\cdot {\left(-2\right)}^3-\frac{1}{9}\cdot {\left(-5\right)}^3$

= $\frac{1}{9}\cdot \left(-8\right)-\frac{1}{9}\cdot \left(-125\right)$

= $-\frac{8}{9}+\frac{125}{9}$

= $13$

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

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

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

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

= $0-\frac{3}{2}+2$

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

= $-\frac{3}{2}+\frac{4}{2}$

= $\frac{1}{2}$

= ${\left[\frac{1}{9}{\left(3x-7\right)}^3+\frac{5}{2}{x}^2\right]}_0^2$

$=$ $\frac{1}{9}\cdot {\left(3\cdot 2-7\right)}^3+\frac{5}{2}\cdot 2^2-\left(\frac{1}{9}\cdot {\left(3\cdot 0-7\right)}^3+\frac{5}{2}\cdot 0^2\right)$

= $\frac{1}{9}\cdot {\left(6-7\right)}^3+\frac{5}{2}\cdot 4-\left(\frac{1}{9}\cdot {\left(0-7\right)}^3+\frac{5}{2}\cdot 0\right)$

= $\frac{1}{9}\cdot {\left(-1\right)}^3+10-\left(\frac{1}{9}\cdot {\left(-7\right)}^3+0\right)$

= $\frac{1}{9}\cdot \left(-1\right)+10-\left(\frac{1}{9}\cdot \left(-343\right)+0\right)$

= $-\frac{1}{9}+10-\left(-\frac{343}{9}+0\right)$

= $-\frac{1}{9}+\frac{90}{9}-\left(-\frac{343}{9}+0\right)$

= $\frac{89}{9}+\frac{343}{9}$

= $48$