$$\underset{0}{\overset{6}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

Somit gilt:

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

= ${\left[-\frac{5}{2}{x}^2+x\right]}_{-1}^0$

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

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

= $0+0-\left(-\frac{5}{2}-1\right)$

= $0-\left(-\frac{5}{2}-\frac{2}{2}\right)$

= $-1·\left(-\frac{7}{2}\right)$

= $\frac{7}{2}$

= ${\left[-\frac{5}{6}{x}^6+\cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{5}{6}\cdot {\left(\frac{3}{2}\pi \right)}^6+\cos \left(\frac{3}{2}\pi \right)-\left(-\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+\cos \left(\frac{1}{2}\pi \right)\right)$

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

= $-\frac{1215}{128}\cdot {\pi }^6+\frac{5}{384}\cdot {\pi }^6$

= $-\frac{3645}{384}\cdot {\pi }^6+\frac{5}{384}\cdot {\pi }^6$

= $-\frac{455}{48}\cdot {\pi }^6$

= ${\left[-\frac{1}{9}{\left(-3x+4\right)}^{-3}\right]}_3^4$

= ${\left[-\frac{1}{9{\left(-3x+4\right)}^3}\right]}_3^4$

$=$ $-\frac{1}{9{\left(-3\cdot 4+4\right)}^3}+\frac{1}{9{\left(-3\cdot 3+4\right)}^3}$

= $-\frac{1}{9{\left(-12+4\right)}^3}+\frac{1}{9{\left(-9+4\right)}^3}$

= $-\frac{1}{9{\left(-8\right)}^3}+\frac{1}{9{\left(-5\right)}^3}$

= $-\frac{1}{9}\cdot \left(-\frac{1}{512}\right)+\frac{1}{9}\cdot \left(-\frac{1}{125}\right)$

= $\frac{1}{4608}-\frac{1}{1125}$

= $\frac{125}{576000}-\frac{512}{576000}$

= $-\frac{43}{64000}$

= ${\left[\cos \left(x\right)+\frac{7}{2}{e}^{-2x}\right]}_0^{\pi }$

$=$ $\cos \left(\pi \right)+\frac{7}{2}{e}^{-2\cdot \pi }-\left(\cos \left(0\right)+\frac{7}{2}{e}^{-2\cdot \left(0\right)}\right)$

= $-1+\frac{7}{2}{e}^{-2\cdot \pi }-\left(1+\frac{7}{2}{e}^0\right)$

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

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

= $\frac{7}{2}{e}^{-2\cdot \pi }-1-1·\frac{9}{2}$

= $\frac{7}{2}{e}^{-2\cdot \pi }-1-\frac{9}{2}$

= $\frac{7}{2}{e}^{-2\pi }-\frac{11}{2}$

= ${\left[\frac{1}{2}\ln \left(\left|$ 2x-1$\right|\right)\right]}_1^4$

$=$ $\frac{1}{2}\ln \left(\left|$ 2\cdot 4-1$\right|\right)-\frac{1}{2}\ln \left(\left|$ 2\cdot 1-1$\right|\right)$

= $\frac{1}{2}\ln \left(\left|$ 8-1$\right|\right)-\frac{1}{2}\ln \left(\left|$ 2-1$\right|\right)$

= $\frac{1}{2}\ln \left(7\right)-\frac{1}{2}\ln \left(\left|$ 2-1$\right|\right)$

= $\frac{1}{2}\ln \left(7\right)-\frac{1}{2}\ln \left(1\right)$

= $\frac{1}{2}\ln \left(7\right)+0$

= $\frac{1}{2}\ln \left(7\right)$