$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ 4 = 2 ⋅ 4 = 8.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

Somit gilt:

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

= ${\left[\frac{5}{3}{x}^3+\frac{1}{2}{x}^2\right]}_{-3}^{-1}$

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

= $\frac{5}{3}\cdot \left(-1\right)+\frac{1}{2}\cdot 1-\left(\frac{5}{3}\cdot \left(-27\right)+\frac{1}{2}\cdot 9\right)$

= $-\frac{5}{3}+\frac{1}{2}-\left(-45+\frac{9}{2}\right)$

= $-\frac{10}{6}+\frac{3}{6}-\left(-\frac{90}{2}+\frac{9}{2}\right)$

= $-\frac{7}{6}-1·\left(-\frac{81}{2}\right)$

= $-\frac{7}{6}+\frac{81}{2}$

= $-\frac{7}{6}+\frac{243}{6}$

= $\frac{118}{3}$

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

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

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

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

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

= $\frac{3}{2}-\frac{8}{9{\pi }^2}+\frac{2}{{\pi }^2}$

= $\frac{3}{2}+\frac{10}{9{\pi }^2}$

= ${\left[-e^{x-2}\right]}_0^2$

$=$ $-e^{2-2}+e^{0-2}$

= $-e^0+e^{-2}$

= $-1+e^{-2}$

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

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

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

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

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

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

= $\frac{1}{3}{e}^{-\frac{3}{2}\pi }+\frac{5}{3}$

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

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

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

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

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

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

= $\frac{1}{2}+\frac{6}{2}-\left(-\frac{1}{2}+0\right)$

= $\frac{7}{2}+\frac{1}{2}$

= $4$