$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 2 = 2 ⋅ 2 = 4.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₅ = (10 - 8) ⋅ $\frac{\mathrm{2\; +\; <mn>1</mn>}}{2}$ = 2 ⋅ 1.5 = 3.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ + I₅ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -8 -4 +2 +4 +3 = -3

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

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

= $-\frac{2}{3}\cdot \left(-8\right)+\frac{3}{2}\cdot 4-\left(-\frac{2}{3}\cdot \left(-27\right)+\frac{3}{2}\cdot 9\right)$

= $\frac{16}{3}+6-\left(18+\frac{27}{2}\right)$

= $\frac{16}{3}+\frac{18}{3}-\left(\frac{36}{2}+\frac{27}{2}\right)$

= $\frac{34}{3}-1·\frac{63}{2}$

= $\frac{34}{3}-\frac{63}{2}$

= $\frac{68}{6}-\frac{189}{6}$

= $-\frac{121}{6}$

= ${\left[-\frac{1}{4}{x}^6+\frac{3}{2}{x}^{-2}\right]}_1^2$

= ${\left[-\frac{1}{4}{x}^6+\frac{3}{2x^2}\right]}_1^2$

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

= $-\frac{1}{4}\cdot 64+\frac{3}{2}\cdot \left(\frac{1}{4}\right)-\left(-\frac{1}{4}\cdot 1+\frac{3}{2}\cdot 1\right)$

= $-16+\frac{3}{8}-\left(-\frac{1}{4}+\frac{3}{2}\right)$

= $-\frac{128}{8}+\frac{3}{8}-\left(-\frac{1}{4}+\frac{6}{4}\right)$

= $-\frac{125}{8}-1·\frac{5}{4}$

= $-\frac{125}{8}-\frac{5}{4}$

= $-\frac{125}{8}-\frac{10}{8}$

= $-\frac{135}{8}$

= ${\left[\frac{2}{3}{e}^{3x-3}\right]}_1^3$

$=$ $\frac{2}{3}{e}^{3\cdot 3-3}-\frac{2}{3}{e}^{3\cdot 1-3}$

= $\frac{2}{3}{e}^{9-3}-\frac{2}{3}{e}^{3-3}$

= $\frac{2}{3}{e}^6-\frac{2}{3}{e}^0$

= $\frac{2}{3}{e}^6-\frac{2}{3}$

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

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

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

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

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

= $-1·\left(-\frac{1}{3}\right)-1·\frac{1}{8}\cdot {\pi }^2+\frac{1}{2}\cdot {\pi }^2$

= $\frac{1}{3}-\frac{1}{8}\cdot {\pi }^2+\frac{1}{2}\cdot {\pi }^2$

= $\frac{1}{3}+\frac{3}{8}\cdot {\pi }^2$

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

$=$ $-3e^{2-2}+3e^{0-2}$

= $-3e^0+3e^{-2}$

= $-3+3e^{-2}$