$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>5</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 4.5 ) = -9.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4 -6 -12 -9 = -23

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

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

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

= $-\frac{16}{3}+4-\left(\frac{2}{3}-2\right)$

= $-\frac{16}{3}+\frac{12}{3}-\left(\frac{2}{3}-\frac{6}{3}\right)$

= $-\frac{4}{3}-1·\left(-\frac{4}{3}\right)$

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

= 0

= ${\left[-\frac{6}{5}{x}^{\frac{5}{4}}+\frac{5}{16}{x}^4\right]}_0^1$

= ${\left[-\frac{6}{5}{\left(\sqrt[4]{x}\right)}^5+\frac{5}{16}{x}^4\right]}_0^1$

$=$ $-\frac{6}{5}{\left(\sqrt[4]{1}\right)}^5+\frac{5}{16}\cdot 1^4-\left(-\frac{6}{5}{\left(\sqrt[4]{0}\right)}^5+\frac{5}{16}\cdot 0^4\right)$

= $-\frac{6}{5}\cdot 1^5+\frac{5}{16}\cdot 1-\left(-\frac{6}{5}\cdot 0^5+\frac{5}{16}\cdot 0\right)$

= $-\frac{6}{5}\cdot 1+\frac{5}{16}-\left(-\frac{6}{5}\cdot 0+0\right)$

= $-\frac{6}{5}+\frac{5}{16}-\left(0+0\right)$

= $-\frac{96}{80}+\frac{25}{80}+0$

= $-\frac{71}{80}$

= ${\left[-e^{-2x+1}\right]}_2^5$

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

= $-e^{-10+1}+e^{-4+1}$

= $-e^{-9}+e^{-3}$

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

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

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

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

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

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

= $5e^{-\frac{3}{2}\pi }+\frac{1}{3}-5e^{-\frac{1}{2}\pi }+\frac{1}{3}$

= $-5e^{-\frac{1}{2}\pi }+5e^{-\frac{3}{2}\pi }+\frac{1}{3}+\frac{1}{3}$

= $-5e^{-\frac{1}{2}\pi }+5e^{-\frac{3}{2}\pi }+\frac{2}{3}$

= ${\left[-{\left(-2x+3\right)}^{-1}\right]}_3^5$

= ${\left[-\frac{1}{-2x+3}\right]}_3^5$

$=$ $-\frac{1}{-2\cdot 5+3}+\frac{1}{-2\cdot 3+3}$

= $-\frac{1}{-10+3}+\frac{1}{-6+3}$

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

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

= $\frac{1}{7}-\frac{1}{3}$

= $\frac{3}{21}-\frac{7}{21}$

= $-\frac{4}{21}$