$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-2\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4 -3 -4 -5 = -8

= ${\left[\frac{1}{3}{x}^3+\frac{5}{2}{x}^2\right]}_0^2$

$=$ $\frac{1}{3}\cdot 2^3+\frac{5}{2}\cdot 2^2-\left(\frac{1}{3}\cdot 0^3+\frac{5}{2}\cdot 0^2\right)$

= $\frac{1}{3}\cdot 8+\frac{5}{2}\cdot 4-\left(\frac{1}{3}\cdot 0+\frac{5}{2}\cdot 0\right)$

= $\frac{8}{3}+10-\left(0+0\right)$

= $\frac{8}{3}+\frac{30}{3}+0$

= $\frac{38}{3}$

= ${\left[5\cdot \cos \left(x\right)-2x^3\right]}_0^{\frac{1}{2}\pi }$

$=$ $5\cdot \cos \left(\frac{1}{2}\pi \right)-2\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(5\cdot \cos \left(0\right)-2\cdot {\left(0\right)}^3\right)$

= $5\cdot 0-2\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(5\cdot 1-2\cdot 0\right)$

= $0-2\cdot {\left(\frac{1}{2}\pi \right)}^3-\left(5+0\right)$

= $-\frac{1}{4}\cdot {\pi }^3-5$

= ${\left[-\frac{2}{9}{\left(3x-7\right)}^3+\frac{5}{2}{x}^2\right]}_0^3$

$=$ $-\frac{2}{9}\cdot {\left(3\cdot 3-7\right)}^3+\frac{5}{2}\cdot 3^2-\left(-\frac{2}{9}\cdot {\left(3\cdot 0-7\right)}^3+\frac{5}{2}\cdot 0^2\right)$

= $-\frac{2}{9}\cdot {\left(9-7\right)}^3+\frac{5}{2}\cdot 9-\left(-\frac{2}{9}\cdot {\left(0-7\right)}^3+\frac{5}{2}\cdot 0\right)$

= $-\frac{2}{9}\cdot 2^3+\frac{45}{2}-\left(-\frac{2}{9}\cdot {\left(-7\right)}^3+0\right)$

= $-\frac{2}{9}\cdot 8+\frac{45}{2}-\left(-\frac{2}{9}\cdot \left(-343\right)+0\right)$

= $-\frac{16}{9}+\frac{45}{2}-\left(\frac{686}{9}+0\right)$

= $-\frac{32}{18}+\frac{405}{18}-\left(\frac{686}{9}+0\right)$

= $\frac{373}{18}-\frac{686}{9}$

= $\frac{373}{18}-\frac{1372}{18}$

= $\frac{373}{18}-\frac{686}{9}$

= $\frac{373}{18}-\frac{1372}{18}$

= $-\frac{111}{2}$

= ${\left[5\cdot \sin \left(x\right)-\frac{1}{3}{x}^{-3}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

= ${\left[5\cdot \sin \left(x\right)-\frac{1}{3x^3}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $-5-\frac{8}{81{\pi }^3}-\left(5-\frac{8}{3{\pi }^3}\right)$

= $-5-\frac{8}{81{\pi }^3}-1·5-1·\left(-\frac{8}{3{\pi }^3}\right)$

= $-5-\frac{8}{81{\pi }^3}-5+\frac{8}{3{\pi }^3}$

= $-5-5-\frac{8}{81{\pi }^3}+\frac{8}{3{\pi }^3}$

= $-10+\frac{208}{81{\pi }^3}$

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

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

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