$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ 4 = 3 ⋅ 4 = 12.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 12 +4 -4.5 -6 = 5.5

= ${\left[-x^2+x\right]}_{-3}^0$

$=$ $-0^2+0-\left(-{\left(-3\right)}^2-3\right)$

= $-0+0-\left(-9-3\right)$

= $-1·\left(-9\right)-1·\left(-3\right)$

= $9+3$

= $12$

= ${\left[-x^3+\frac{2}{5}{x}^{\frac{5}{2}}\right]}_4^9$

= ${\left[-x^3+\frac{2}{5}{\left(\sqrt{x}\right)}^5\right]}_4^9$

$=$ $-9^3+\frac{2}{5}{\left(\sqrt{9}\right)}^5-\left(-4^3+\frac{2}{5}{\left(\sqrt{4}\right)}^5\right)$

= $-729+\frac{2}{5}\cdot 3^5-\left(-64+\frac{2}{5}\cdot 2^5\right)$

= $-729+\frac{2}{5}\cdot 243-\left(-64+\frac{2}{5}\cdot 32\right)$

= $-729+\frac{486}{5}-\left(-64+\frac{64}{5}\right)$

= $-\frac{3645}{5}+\frac{486}{5}-\left(-\frac{320}{5}+\frac{64}{5}\right)$

= $-\frac{3159}{5}-1·\left(-\frac{256}{5}\right)$

= $-\frac{3159}{5}+\frac{256}{5}$

= $-\frac{2903}{5}$

= ${\left[-\frac{4}{3}{\left(x-2\right)}^{\frac{3}{2}}\right]}_{18}^{27}$

= ${\left[-\frac{4}{3}{\left(\sqrt{x-2}\right)}^3\right]}_{18}^{27}$

$=$ $-\frac{4}{3}{\left(\sqrt{27-2}\right)}^3+\frac{4}{3}{\left(\sqrt{18-2}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{25}\right)}^3+\frac{4}{3}{\left(\sqrt{16}\right)}^3$

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

= $-\frac{4}{3}\cdot 125+\frac{4}{3}\cdot 64$

= $-\frac{500}{3}+\frac{256}{3}$

= $-\frac{244}{3}$

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

$=$ $-2\cdot \sin \left(\pi \right)-\frac{1}{4}{e}^{2\cdot \pi }-\left(-2\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{1}{4}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

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

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

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

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

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

= ${\left[-\frac{2}{3}{\left(2x-2\right)}^{\frac{3}{2}}\right]}_3^9$

= ${\left[-\frac{2}{3}{\left(\sqrt{2x-2}\right)}^3\right]}_3^9$

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

= $-\frac{2}{3}{\left(\sqrt{18-2}\right)}^3+\frac{2}{3}{\left(\sqrt{6-2}\right)}^3$

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

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

= $-\frac{2}{3}\cdot 64+\frac{2}{3}\cdot 8$

= $-\frac{128}{3}+\frac{16}{3}$

= $-\frac{112}{3}$