$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 2 = 2 ⋅ 2 = 4.

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

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

Somit gilt:

$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ = 4 +2 -1 = 5

= ${\left[2x^2+2x\right]}_1^3$

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

= $2\cdot 9+6-\left(2\cdot 1+2\right)$

= $18+6-\left(2+2\right)$

= $24-1·4$

= $24-4$

= $20$

= ${\left[\frac{9}{4}\cdot \sin \left(x\right)+\frac{2}{3}{x}^3\right]}_0^{\pi }$

$=$ $\frac{9}{4}\cdot \sin \left(\pi \right)+\frac{2}{3}\cdot {\pi }^3-\left(\frac{9}{4}\cdot \sin \left(0\right)+\frac{2}{3}\cdot {\left(0\right)}^3\right)$

= $\frac{9}{4}\cdot 0+\frac{2}{3}\cdot {\pi }^3-\left(\frac{9}{4}\cdot 0+\frac{2}{3}\cdot 0\right)$

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

= $\frac{2}{3}\cdot {\pi }^3+0$

= $\frac{2}{3}\cdot {\pi }^3$

= ${\left[\frac{2}{3}{e}^{-3x+6}\right]}_0^2$

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

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

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

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

= ${\left[\frac{6}{5}{x}^5-\frac{8}{3}{e}^{3x}\right]}_{-1}^1$

$=$ $\frac{6}{5}\cdot 1^5-\frac{8}{3}{e}^{3\cdot 1}-\left(\frac{6}{5}\cdot {\left(-1\right)}^5-\frac{8}{3}{e}^{3\cdot \left(-1\right)}\right)$

= $\frac{6}{5}\cdot 1-\frac{8}{3}{e}^3-\left(\frac{6}{5}\cdot \left(-1\right)-\frac{8}{3}{e}^{-3}\right)$

= $\frac{6}{5}-\frac{8}{3}{e}^3-\left(-\frac{6}{5}-\frac{8}{3}{e}^{-3}\right)$

= $-\frac{8}{3}{e}^3+\frac{6}{5}-\left(-\frac{8}{3}{e}^{-3}-\frac{6}{5}\right)$

= $-\frac{8}{3}{e}^3+\frac{6}{5}+\frac{8}{3}{e}^{-3}-1·\left(-\frac{6}{5}\right)$

= $-\frac{8}{3}{e}^3+\frac{6}{5}+\frac{8}{3}{e}^{-3}+\frac{6}{5}$

= $-\frac{8}{3}{e}^3+\frac{8}{3}{e}^{-3}+\frac{6}{5}+\frac{6}{5}$

= $-\frac{8}{3}{e}^3+\frac{8}{3}{e}^{-3}+\frac{12}{5}$

= ${\left[\frac{1}{2}{\left(-x+1\right)}^4+\frac{5}{2}{x}^2\right]}_2^5$

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

= $\frac{1}{2}\cdot {\left(-4\right)}^4+\frac{5}{2}\cdot 25-\left(\frac{1}{2}\cdot {\left(-1\right)}^4+\frac{5}{2}\cdot 4\right)$

= $\frac{1}{2}\cdot 256+\frac{125}{2}-\left(\frac{1}{2}\cdot 1+10\right)$

= $128+\frac{125}{2}-\left(\frac{1}{2}+10\right)$

= $\frac{256}{2}+\frac{125}{2}-\left(\frac{1}{2}+\frac{20}{2}\right)$

= $\frac{381}{2}-1·\frac{21}{2}$

= $\frac{381}{2}-\frac{21}{2}$

= $180$