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

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

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

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

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 -2 -4 -6 = -9

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

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

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

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

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

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

= $\frac{1}{3}+\frac{1}{3}$

= $\frac{2}{3}$

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

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

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

= $-\frac{5}{3}\cdot \left(\frac{1}{8}\right)-\frac{5}{2}\cdot 4-\left(-\frac{5}{3}\cdot 1-\frac{5}{2}\cdot 1\right)$

= $-\frac{5}{24}-10-\left(-\frac{5}{3}-\frac{5}{2}\right)$

= $-\frac{5}{24}-\frac{240}{24}-\left(-\frac{10}{6}-\frac{15}{6}\right)$

= $-\frac{245}{24}-1·\left(-\frac{25}{6}\right)$

= $-\frac{245}{24}+\frac{25}{6}$

= $-\frac{245}{24}+\frac{100}{24}$

= $-\frac{145}{24}$

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

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

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

= $\frac{2}{3}\cdot 0-\frac{2}{3}\cdot 1$

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

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

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

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

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

= $-1+0-\left(-0+1\right)$

= $-1-1$

= $-2$

= ${\left[-\frac{1}{6}{\left(3x-4\right)}^4+x\right]}_2^4$

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

= $-\frac{1}{6}\cdot {\left(12-4\right)}^4+4-\left(-\frac{1}{6}\cdot {\left(6-4\right)}^4+2\right)$

= $-\frac{1}{6}\cdot 8^4+4-\left(-\frac{1}{6}\cdot 2^4+2\right)$

= $-\frac{1}{6}\cdot 4096+4-\left(-\frac{1}{6}\cdot 16+2\right)$

= $-\frac{2048}{3}+4-\left(-\frac{8}{3}+2\right)$

= $-\frac{2048}{3}+\frac{12}{3}-\left(-\frac{8}{3}+\frac{6}{3}\right)$

= $-\frac{2036}{3}-1·\left(-\frac{2}{3}\right)$

= $-\frac{2036}{3}+\frac{2}{3}$

= $-678$