$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-1\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 2.5 ) = -7.5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 3 -1.5 -2 -7.5 = -8

= ${\left[2x^2+4x\right]}_{-2}^{-1}$

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

= $2\cdot 1-4-\left(2\cdot 4-8\right)$

= $2-4-\left(8-8\right)$

= $-2-1·0$

= $-2+0$

= $-2$

= ${\left[-\frac{5}{3}{x}^{-3}+\frac{7}{6}{x}^6\right]}_2^3$

= ${\left[-\frac{5}{3x^3}+\frac{7}{6}{x}^6\right]}_2^3$

$=$ $-\frac{5}{3\cdot 3^3}+\frac{7}{6}\cdot 3^6-\left(-\frac{5}{3\cdot 2^3}+\frac{7}{6}\cdot 2^6\right)$

= $-\frac{5}{3}\cdot \left(\frac{1}{27}\right)+\frac{7}{6}\cdot 729-\left(-\frac{5}{3}\cdot \left(\frac{1}{8}\right)+\frac{7}{6}\cdot 64\right)$

= $-\frac{5}{81}+\frac{1701}{2}-\left(-\frac{5}{24}+\frac{224}{3}\right)$

= $-\frac{10}{162}+\frac{137781}{162}-\left(-\frac{5}{24}+\frac{1792}{24}\right)$

= $\frac{137771}{162}-1·\frac{1787}{24}$

= $\frac{137771}{162}-\frac{1787}{24}$

= $\frac{551084}{648}-\frac{48249}{648}$

= $\frac{502835}{648}$

= ${\left[\frac{1}{6}{\left(-3x+7\right)}^4+4x\right]}_0^2$

$=$ $\frac{1}{6}\cdot {\left(-3\cdot 2+7\right)}^4+4\cdot 2-\left(\frac{1}{6}\cdot {\left(-3\cdot 0+7\right)}^4+4\cdot 0\right)$

= $\frac{1}{6}\cdot {\left(-6+7\right)}^4+8-\left(\frac{1}{6}\cdot {\left(0+7\right)}^4+0\right)$

= $\frac{1}{6}\cdot 1^4+8-\left(\frac{1}{6}\cdot 7^4+0\right)$

= $\frac{1}{6}\cdot 1+8-\left(\frac{1}{6}\cdot 2401+0\right)$

= $\frac{1}{6}+8-\left(\frac{2401}{6}+0\right)$

= $\frac{1}{6}+\frac{48}{6}-\left(\frac{2401}{6}+0\right)$

= $\frac{49}{6}-\frac{2401}{6}$

= $-392$

= ${\left[-3\cdot \sin \left(x\right)+\frac{4}{3}{x}^{-3}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

= ${\left[-3\cdot \sin \left(x\right)+\frac{4}{3x^3}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $3+\frac{32}{81{\pi }^3}-\left(-3+\frac{32}{3{\pi }^3}\right)$

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

= $3+\frac{32}{81{\pi }^3}+3-\frac{32}{3{\pi }^3}$

= $3+3+\frac{32}{81{\pi }^3}-\frac{32}{3{\pi }^3}$

= $6-\frac{832}{81{\pi }^3}$

= ${\left[\ln \left(\left|$ -x+2$\right|\right)\right]}_4^5$

$=$ $\ln \left(\left|$ -5+2$\right|\right)-\ln \left(\left|$ -4+2$\right|\right)$

= $\ln \left(\left|$ -5+2$\right|\right)-\ln \left(\left|$ -4+2$\right|\right)$

= $\ln \left(3\right)-\ln \left(\left|$ -4+2$\right|\right)$

= $\ln \left(3\right)-\ln \left(2\right)$