$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 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.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{-1\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

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

= ${\left[\frac{5}{2}{x}^2+x\right]}_0^3$

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

= $\frac{5}{2}\cdot 9+3-\left(\frac{5}{2}\cdot 0+0\right)$

= $\frac{45}{2}+3-\left(0+0\right)$

= $\frac{45}{2}+\frac{6}{2}+0$

= $\frac{51}{2}$

= ${\left[-\frac{10}{3}{x}^{\frac{3}{2}}+\frac{2}{3}{e}^x\right]}_0^4$

= ${\left[-\frac{10}{3}{\left(\sqrt{x}\right)}^3+\frac{2}{3}{e}^x\right]}_0^4$

$=$ $-\frac{10}{3}{\left(\sqrt{4}\right)}^3+\frac{2}{3}{e}^4-\left(-\frac{10}{3}{\left(\sqrt{0}\right)}^3+\frac{2}{3}{e}^0\right)$

= $-\frac{10}{3}\cdot 2^3+\frac{2}{3}{e}^4-\left(-\frac{10}{3}\cdot 0^3+\frac{2}{3}\right)$

= $-\frac{10}{3}\cdot 8+\frac{2}{3}{e}^4-\left(-\frac{10}{3}\cdot 0+\frac{2}{3}\right)$

= $-\frac{80}{3}+\frac{2}{3}{e}^4-\left(0+\frac{2}{3}\right)$

= $\frac{2}{3}{e}^4-\frac{80}{3}-\left(0+\frac{2}{3}\right)$

= $\frac{2}{3}{e}^4-\frac{80}{3}-\frac{2}{3}$

= $\frac{2}{3}{e}^4-\frac{82}{3}$

= ${\left[\frac{1}{3}\cdot \sin \left(-3x+\frac{1}{2}\pi \right)\right]}_0^{\frac{1}{2}\pi }$

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

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

= $\frac{1}{3}\cdot 0-\frac{1}{3}\cdot 1$

= $0-\frac{1}{3}$

= $0-\frac{1}{3}$

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

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

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

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

= $0-4-\left(0+4\right)$

= $-4-4$

= $-8$

= ${\left[-\frac{1}{6}{\left(3x-3\right)}^4-\frac{1}{2}{x}^2\right]}_2^4$

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

= $-\frac{1}{6}\cdot {\left(12-3\right)}^4-\frac{1}{2}\cdot 16-\left(-\frac{1}{6}\cdot {\left(6-3\right)}^4-\frac{1}{2}\cdot 4\right)$

= $-\frac{1}{6}\cdot 9^4-8-\left(-\frac{1}{6}\cdot 3^4-2\right)$

= $-\frac{1}{6}\cdot 6561-8-\left(-\frac{1}{6}\cdot 81-2\right)$

= $-\frac{2187}{2}-8-\left(-\frac{27}{2}-2\right)$

= $-\frac{2187}{2}-\frac{16}{2}-\left(-\frac{27}{2}-\frac{4}{2}\right)$

= $-\frac{2203}{2}-1·\left(-\frac{31}{2}\right)$

= $-\frac{2203}{2}+\frac{31}{2}$

= $-1086$