$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

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

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4.5 -2 -4 -7.5 = -9

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

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

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

= $2+1-\left(8+2\right)$

= $3-1·10$

= $3-10$

= $-7$

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

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

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

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

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

= $-\frac{8}{81{\pi }^3}+\frac{8}{3{\pi }^3}$

= $-\frac{8}{81{\pi }^3}+\frac{216}{81{\pi }^3}$

= $\frac{208}{81{\pi }^3}$

= ${\left[-\frac{1}{2}{\left(-2x+4\right)}^{-1}\right]}_4^7$

= ${\left[-\frac{1}{2\left(-2x+4\right)}\right]}_4^7$

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

= $-\frac{1}{2\left(-14+4\right)}+\frac{1}{2\left(-8+4\right)}$

= $-\frac{1}{2\left(-10\right)}+\frac{1}{2\left(-4\right)}$

= $-\frac{1}{2}\cdot \left(-\frac{1}{10}\right)+\frac{1}{2}\cdot \left(-\frac{1}{4}\right)$

= $\frac{1}{20}-\frac{1}{8}$

= $\frac{2}{40}-\frac{5}{40}$

= $-\frac{3}{40}$

= ${\left[-\frac{2}{5}{x}^5-4\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{2}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5-4\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{2}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5-4\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{2}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5-4\cdot \left(-1\right)-\left(-\frac{2}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5-4\cdot 1\right)$

= $-\frac{2}{5}\cdot {\left(\frac{3}{2}\pi \right)}^5+4-\left(-\frac{2}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5-4\right)$

= $4-\frac{243}{80}\cdot {\pi }^5-\left(-4-\frac{1}{80}\cdot {\pi }^5\right)$

= $4-\frac{243}{80}\cdot {\pi }^5-1·\left(-4\right)-1·\left(-\frac{1}{80}\cdot {\pi }^5\right)$

= $4-\frac{243}{80}\cdot {\pi }^5+4+\frac{1}{80}\cdot {\pi }^5$

= $4+4-\frac{243}{80}\cdot {\pi }^5+\frac{1}{80}\cdot {\pi }^5$

= $8-\frac{121}{40}\cdot {\pi }^5$

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

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

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

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

= $\frac{1}{3}+\frac{1}{3}$

= $\frac{2}{3}$