$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 1 ) = 3 ⋅ ( - 1 ) = -3.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 2 -1.5 -3 = -2.5

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

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

= $-\frac{2}{3}\cdot 0+0-\left(-\frac{2}{3}\cdot \left(-8\right)+10\right)$

= $0+0-\left(\frac{16}{3}+10\right)$

= $0-\left(\frac{16}{3}+\frac{30}{3}\right)$

= $-1·\frac{46}{3}$

= $-\frac{46}{3}$

= ${\left[\frac{1}{6}{x}^{-2}+\frac{4}{5}{x}^5\right]}_2^6$

= ${\left[\frac{1}{6x^2}+\frac{4}{5}{x}^5\right]}_2^6$

$=$ $\frac{1}{6\cdot 6^2}+\frac{4}{5}\cdot 6^5-\left(\frac{1}{6\cdot 2^2}+\frac{4}{5}\cdot 2^5\right)$

= $\frac{1}{6}\cdot \left(\frac{1}{36}\right)+\frac{4}{5}\cdot 7776-\left(\frac{1}{6}\cdot \left(\frac{1}{4}\right)+\frac{4}{5}\cdot 32\right)$

= $\frac{1}{216}+\frac{31104}{5}-\left(\frac{1}{24}+\frac{128}{5}\right)$

= $\frac{5}{1080}+\frac{6718464}{1080}-\left(\frac{5}{120}+\frac{3072}{120}\right)$

= $\frac{6718469}{1080}-1·\frac{3077}{120}$

= $\frac{6718469}{1080}-\frac{3077}{120}$

= $\frac{6718469}{1080}-\frac{27693}{1080}$

= $\frac{836347}{135}$

= ${\left[\frac{2}{3}{\left(x-3\right)}^3+6x\right]}_2^5$

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

= $\frac{2}{3}\cdot 2^3+30-\left(\frac{2}{3}\cdot {\left(-1\right)}^3+12\right)$

= $\frac{2}{3}\cdot 8+30-\left(\frac{2}{3}\cdot \left(-1\right)+12\right)$

= $\frac{16}{3}+30-\left(-\frac{2}{3}+12\right)$

= $\frac{16}{3}+\frac{90}{3}-\left(-\frac{2}{3}+\frac{36}{3}\right)$

= $\frac{106}{3}-1·\frac{34}{3}$

= $\frac{106}{3}-\frac{34}{3}$

= $24$

= ${\left[5\cdot \sin \left(x\right)-5\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

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

= $5\cdot 1-5\cdot 0-\left(5\cdot 0-5\cdot 1\right)$

= $5+0-\left(0-5\right)$

= $5+5$

= $10$

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

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

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

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

= $-\frac{2}{9}\cdot 125-3-\left(-\frac{2}{9}\cdot \left(-1\right)-1\right)$

= $-\frac{250}{9}-3-\left(\frac{2}{9}-1\right)$

= $-\frac{250}{9}-\frac{27}{9}-\left(\frac{2}{9}-\frac{9}{9}\right)$

= $-\frac{277}{9}-1·\left(-\frac{7}{9}\right)$

= $-\frac{277}{9}+\frac{7}{9}$

= $-30$