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

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

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 +2 +4 = 3

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

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

= $-\frac{1}{3}\cdot 1+2-\left(-\frac{1}{3}\cdot \left(-27\right)-6\right)$

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

= $-\frac{1}{3}+\frac{6}{3}-1·3$

= $\frac{5}{3}-3$

= $\frac{5}{3}-\frac{9}{3}$

= $-\frac{4}{3}$

= ${\left[-2x^{\frac{3}{2}}+\frac{1}{12}{x}^4\right]}_1^9$

= ${\left[-2{\left(\sqrt{x}\right)}^3+\frac{1}{12}{x}^4\right]}_1^9$

$=$ $-2{\left(\sqrt{9}\right)}^3+\frac{1}{12}\cdot 9^4-\left(-2{\left(\sqrt{1}\right)}^3+\frac{1}{12}\cdot 1^4\right)$

= $-2\cdot 3^3+\frac{1}{12}\cdot 6561-\left(-2\cdot 1^3+\frac{1}{12}\cdot 1\right)$

= $-2\cdot 27+\frac{2187}{4}-\left(-2\cdot 1+\frac{1}{12}\right)$

= $-54+\frac{2187}{4}-\left(-2+\frac{1}{12}\right)$

= $-\frac{216}{4}+\frac{2187}{4}-\left(-\frac{24}{12}+\frac{1}{12}\right)$

= $\frac{1971}{4}-1·\left(-\frac{23}{12}\right)$

= $\frac{1971}{4}+\frac{23}{12}$

= $\frac{5913}{12}+\frac{23}{12}$

= $\frac{1971}{4}+\frac{23}{12}$

= $\frac{5913}{12}+\frac{23}{12}$

= $\frac{1484}{3}$

= ${\left[-\frac{3}{8}{\left(2x-5\right)}^4+2x\right]}_2^3$

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

= $-\frac{3}{8}\cdot {\left(6-5\right)}^4+6-\left(-\frac{3}{8}\cdot {\left(4-5\right)}^4+4\right)$

= $-\frac{3}{8}\cdot 1^4+6-\left(-\frac{3}{8}\cdot {\left(-1\right)}^4+4\right)$

= $-\frac{3}{8}\cdot 1+6-\left(-\frac{3}{8}\cdot 1+4\right)$

= $-\frac{3}{8}+6-\left(-\frac{3}{8}+4\right)$

= $-\frac{3}{8}+\frac{48}{8}-\left(-\frac{3}{8}+\frac{32}{8}\right)$

= $\frac{45}{8}-1·\frac{29}{8}$

= $\frac{45}{8}-\frac{29}{8}$

= $2$

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

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

= $3\cdot 0-2\cdot \left(-1\right)-\left(3\cdot 1-2\cdot 0\right)$

= $0+2-\left(3+0\right)$

= $2-3$

= $-1$

= ${\left[-\frac{3}{2}\ln \left(\left|$ 2x-1$\right|\right)\right]}_1^4$

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

= $-\frac{3}{2}\ln \left(\left|$ 8-1$\right|\right)+\frac{3}{2}\ln \left(\left|$ 2-1$\right|\right)$

= $-\frac{3}{2}\ln \left(7\right)+\frac{3}{2}\ln \left(\left|$ 2-1$\right|\right)$

= $-\frac{3}{2}\ln \left(7\right)+\frac{3}{2}\ln \left(1\right)$

= $-\frac{3}{2}\ln \left(7\right)+0$

= $-\frac{3}{2}\ln \left(7\right)$