$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

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

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 2 = 2 ⋅ 2 = 4.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{2\; +\; <mn>3</mn>}}{2}$ = 2 ⋅ 2.5 = 5.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₅ = (10 - 8) ⋅ 3 = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ + I₅ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 +2 +4 +5 +6 = 14

= ${\left[\frac{5}{3}{x}^3+4x\right]}_{-3}^{-2}$

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

= $\frac{5}{3}\cdot \left(-8\right)-8-\left(\frac{5}{3}\cdot \left(-27\right)-12\right)$

= $-\frac{40}{3}-8-\left(-45-12\right)$

= $-\frac{40}{3}-\frac{24}{3}-1·\left(-57\right)$

= $-\frac{64}{3}+57$

= $-\frac{64}{3}+\frac{171}{3}$

= $\frac{107}{3}$

= ${\left[4\cdot \sin \left(x\right)+\frac{5}{3}{x}^3\right]}_0^{\pi }$

$=$ $4\cdot \sin \left(\pi \right)+\frac{5}{3}\cdot {\pi }^3-\left(4\cdot \sin \left(0\right)+\frac{5}{3}\cdot {\left(0\right)}^3\right)$

= $4\cdot 0+\frac{5}{3}\cdot {\pi }^3-\left(4\cdot 0+\frac{5}{3}\cdot 0\right)$

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

= $\frac{5}{3}\cdot {\pi }^3+0$

= $\frac{5}{3}\cdot {\pi }^3$

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

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

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

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

= ${\left[-3\cdot \cos \left(x\right)+\frac{7}{2}\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $-3\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{7}{2}\cdot \sin \left(\frac{1}{2}\pi \right)-\left(-3\cdot \cos \left(0\right)+\frac{7}{2}\cdot \sin \left(0\right)\right)$

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

= $0+\frac{7}{2}-\left(-3+0\right)$

= $0+\frac{7}{2}+3$

= $\frac{7}{2}+3$

= $\frac{7}{2}+\frac{6}{2}$

= $\frac{13}{2}$

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

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

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

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

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

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

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