$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₄ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 6) ⋅ 4 = 3 ⋅ 4 = 12.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -6 -3 +4 +12 = 7

= ${\left[-\frac{5}{3}{x}^3+x^2\right]}_{-2}^1$

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

= $-\frac{5}{3}\cdot 1+1-\left(-\frac{5}{3}\cdot \left(-8\right)+4\right)$

= $-\frac{5}{3}+1-\left(\frac{40}{3}+4\right)$

= $-\frac{5}{3}+\frac{3}{3}-\left(\frac{40}{3}+\frac{12}{3}\right)$

= $-\frac{2}{3}-1·\frac{52}{3}$

= $-\frac{2}{3}-\frac{52}{3}$

= $-18$

= ${\left[\frac{10}{3}{x}^{\frac{3}{2}}+\frac{1}{3}\cdot \sin \left(x\right)\right]}_4^9$

= ${\left[\frac{10}{3}{\left(\sqrt{x}\right)}^3+\frac{1}{3}\cdot \sin \left(x\right)\right]}_4^9$

$=$ $\frac{10}{3}{\left(\sqrt{9}\right)}^3+\frac{1}{3}\cdot \sin \left(9\right)-\left(\frac{10}{3}{\left(\sqrt{4}\right)}^3+\frac{1}{3}\cdot \sin \left(4\right)\right)$

= $\frac{10}{3}\cdot 3^3+\frac{1}{3}\cdot \sin \left(9\right)-\left(\frac{10}{3}\cdot 2^3+\frac{1}{3}\cdot \sin \left(4\right)\right)$

= $\frac{10}{3}\cdot 27+\frac{1}{3}\cdot \sin \left(9\right)-\left(\frac{10}{3}\cdot 8+\frac{1}{3}\cdot \sin \left(4\right)\right)$

= $90+\frac{1}{3}\cdot \sin \left(9\right)-\left(\frac{80}{3}+\frac{1}{3}\cdot \sin \left(4\right)\right)$

= $\frac{1}{3}\cdot \sin \left(9\right)+90-\left(\frac{1}{3}\cdot \sin \left(4\right)+\frac{80}{3}\right)$

= $\frac{1}{3}\cdot \sin \left(9\right)+90-1·\frac{1}{3}\cdot \sin \left(4\right)-1·\frac{80}{3}$

= $\frac{1}{3}\cdot \sin \left(9\right)+90-\frac{1}{3}\cdot \sin \left(4\right)-\frac{80}{3}$

= $\frac{1}{3}\cdot \sin \left(9\right)-\frac{1}{3}\cdot \sin \left(4\right)+90-\frac{80}{3}$

= $\frac{1}{3}\cdot \sin \left(9\right)-\frac{1}{3}\cdot \sin \left(4\right)+\frac{190}{3}$

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

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

= $e^{-8+1}-e^{-2+1}$

= $e^{-7}-e^{-1}$

= ${\left[6\cdot \cos \left(x\right)-\frac{4}{3}{x}^3\right]}_0^{\frac{1}{2}\pi }$

$=$ $6\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{4}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(6\cdot \cos \left(0\right)-\frac{4}{3}\cdot {\left(0\right)}^3\right)$

= $6\cdot 0-\frac{4}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(6\cdot 1-\frac{4}{3}\cdot 0\right)$

= $0-\frac{4}{3}\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(6+0\right)$

= $-\frac{1}{6}\cdot {\pi }^3-6$

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

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

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

= $-\frac{1}{6}\cdot {\left(-4\right)}^3+\frac{1}{6}\cdot 0^3$

= $-\frac{1}{6}\cdot \left(-64\right)+\frac{1}{6}\cdot 0$

= $\frac{32}{3}+0$

= $\frac{32}{3}+0$

= $\frac{32}{3}$