$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -6 -4.5 +1.5 +2 = -7

= ${\left[-\frac{3}{2}{x}^2+5x\right]}_1^4$

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

= $-\frac{3}{2}\cdot 16+20-\left(-\frac{3}{2}\cdot 1+5\right)$

= $-24+20-\left(-\frac{3}{2}+5\right)$

= $-4-\left(-\frac{3}{2}+\frac{10}{2}\right)$

= $-4-1·\frac{7}{2}$

= $-4-\frac{7}{2}$

= $-\frac{8}{2}-\frac{7}{2}$

= $-\frac{15}{2}$

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

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

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

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

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

= $\frac{7}{8{\pi }^2}-\left(-2+\frac{14}{{\pi }^2}\right)$

= $-1·\left(-2\right)-1·\frac{14}{{\pi }^2}+\frac{7}{8{\pi }^2}$

= $2-\frac{14}{{\pi }^2}+\frac{7}{8{\pi }^2}$

= $2-\frac{105}{8{\pi }^2}$

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

$=$ $-e^{-3+3}+e^{-1+3}$

= $-e^0+e^2$

= $-1+e^2$

= ${\left[2\cdot \cos \left(x\right)-\frac{2}{9}{x}^6\right]}_0^{\frac{1}{2}\pi }$

$=$ $2\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6-\left(2\cdot \cos \left(0\right)-\frac{2}{9}\cdot {\left(0\right)}^6\right)$

= $2\cdot 0-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6-\left(2\cdot 1-\frac{2}{9}\cdot 0\right)$

= $0-\frac{2}{9}\cdot {\left(\frac{1}{2}\pi \right)}^6-\left(2+0\right)$

= $-\frac{1}{288}\cdot {\pi }^6-2$

= ${\left[-\frac{1}{3}{\left(2x-2\right)}^{\frac{3}{2}}\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

= ${\left[-\frac{1}{3}{\left(\sqrt{2x-2}\right)}^3\right]}_{\frac{11}{2}}^{\frac{27}{2}}$

$=$ $-\frac{1}{3}{\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3+\frac{1}{3}{\left(\sqrt{2\cdot \left(\frac{11}{2}\right)-2}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{27-2}\right)}^3+\frac{1}{3}{\left(\sqrt{11-2}\right)}^3$

= $-\frac{1}{3}{\left(\sqrt{25}\right)}^3+\frac{1}{3}{\left(\sqrt{9}\right)}^3$

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

= $-\frac{1}{3}\cdot 125+\frac{1}{3}\cdot 27$

= $-\frac{125}{3}+9$

= $-\frac{125}{3}+\frac{27}{3}$

= $-\frac{98}{3}$