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

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ 2 = 3 ⋅ 2 = 6.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -4.5 +3 +6 = 4.5

= ${\left[-\frac{1}{3}{x}^3-3x\right]}_{-1}^1$

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

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

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

= $-\frac{1}{3}-\frac{9}{3}-\left(\frac{1}{3}+\frac{9}{3}\right)$

= $-\frac{10}{3}-1·\frac{10}{3}$

= $-\frac{10}{3}-\frac{10}{3}$

= $-\frac{20}{3}$

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

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

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

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

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

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

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

= $3+\frac{1}{4{\pi }^2}+3-\frac{1}{{\pi }^2}$

= $3+3+\frac{1}{4{\pi }^2}-\frac{1}{{\pi }^2}$

= $6-\frac{3}{4{\pi }^2}$

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

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

= $\frac{2}{3}\cdot 2^3-16-\left(\frac{2}{3}\cdot 0^3-4\right)$

= $\frac{2}{3}\cdot 8-16-\left(\frac{2}{3}\cdot 0-4\right)$

= $\frac{16}{3}-16-\left(0-4\right)$

= $\frac{16}{3}-\frac{48}{3}+4$

= $-\frac{32}{3}+4$

= $-\frac{32}{3}+\frac{12}{3}$

= $-\frac{20}{3}$

= ${\left[-\frac{7}{3}\cdot \sin \left(x\right)+4\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{7}{3}\cdot \sin \left(\pi \right)+4\cdot \cos \left(\pi \right)-\left(-\frac{7}{3}\cdot \sin \left(\frac{1}{2}\pi \right)+4\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

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

= $0-4-\left(-\frac{7}{3}+0\right)$

= $-4-\left(-\frac{7}{3}+0\right)$

= $-4+\frac{7}{3}$

= $-\frac{12}{3}+\frac{7}{3}$

= $-\frac{5}{3}$

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

$=$ $-\frac{1}{2}{e}^{2\cdot 4-3}+\frac{1}{2}{e}^{2\cdot 2-3}$

= $-\frac{1}{2}{e}^{8-3}+\frac{1}{2}{e}^{4-3}$

= $-\frac{1}{2}{e}^5+\frac{1}{2}e$

= $-\frac{1}{2}{e}^5+\frac{1}{2}e$