$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

Somit gilt:

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

= ${\left[\frac{5}{3}{x}^3+x^2\right]}_{-3}^0$

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

= $\frac{5}{3}\cdot 0+0-\left(\frac{5}{3}\cdot \left(-27\right)+9\right)$

= $0+0-\left(-45+9\right)$

= $0-1·\left(-36\right)$

= $36$

= ${\left[\frac{2}{3}{x}^3+2x^{-1}\right]}_1^3$

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

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

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

= $18+\frac{2}{3}-\left(\frac{2}{3}+2\right)$

= $\frac{54}{3}+\frac{2}{3}-\left(\frac{2}{3}+\frac{6}{3}\right)$

= $\frac{56}{3}-1·\frac{8}{3}$

= $\frac{56}{3}-\frac{8}{3}$

= $16$

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

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

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

= $0-1$

= $-1$

= ${\left[-\frac{7}{3}\cdot \sin \left(x\right)+\frac{2}{9}{x}^6\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $\frac{7}{3}+\frac{81}{32}\cdot {\pi }^6-\left(-\frac{7}{3}+\frac{1}{288}\cdot {\pi }^6\right)$

= $\frac{7}{3}+\frac{81}{32}\cdot {\pi }^6-1·\left(-\frac{7}{3}\right)-1·\frac{1}{288}\cdot {\pi }^6$

= $\frac{7}{3}+\frac{81}{32}\cdot {\pi }^6+\frac{7}{3}-\frac{1}{288}\cdot {\pi }^6$

= $\frac{7}{3}+\frac{7}{3}+\frac{81}{32}\cdot {\pi }^6-\frac{1}{288}\cdot {\pi }^6$

= $\frac{14}{3}+\frac{91}{36}\cdot {\pi }^6$

= ${\left[3e^{x-1}\right]}_1^2$

$=$ $3e^{2-1}-3e^{1-1}$

= $3e-3e^0$

= $3e-3$