$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

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 = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 7) ⋅ 3 = 3 ⋅ 3 = 9.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4.5 +3 +9 = 7.5

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

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

= $\frac{3}{2}\cdot 9-15-\left(\frac{3}{2}\cdot 0+0\right)$

= $\frac{27}{2}-15-\left(0+0\right)$

= $\frac{27}{2}-\frac{30}{2}+0$

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

= ${\left[-\frac{4}{3}{x}^{\frac{3}{2}}+\frac{5}{4}\cdot \sin \left(x\right)\right]}_0^{16}$

= ${\left[-\frac{4}{3}{\left(\sqrt{x}\right)}^3+\frac{5}{4}\cdot \sin \left(x\right)\right]}_0^{16}$

$=$ $-\frac{4}{3}{\left(\sqrt{16}\right)}^3+\frac{5}{4}\cdot \sin \left(16\right)-\left(-\frac{4}{3}{\left(\sqrt{0}\right)}^3+\frac{5}{4}\cdot \sin \left(0\right)\right)$

= $-\frac{4}{3}\cdot 4^3+\frac{5}{4}\cdot \sin \left(16\right)-\left(-\frac{4}{3}\cdot 0^3+\frac{5}{4}\cdot 0\right)$

= $-\frac{4}{3}\cdot 64+\frac{5}{4}\cdot \sin \left(16\right)-\left(-\frac{4}{3}\cdot 0+0\right)$

= $-\frac{256}{3}+\frac{5}{4}\cdot \sin \left(16\right)-\left(0+0\right)$

= $\frac{5}{4}\cdot \sin \left(16\right)-\frac{256}{3}+0$

= $\frac{5}{4}\cdot \sin \left(16\right)-\frac{256}{3}$

= ${\left[-\frac{2}{3}{\left(-2x+1\right)}^{\frac{3}{2}}\right]}_{-12}^{-\frac{15}{2}}$

= ${\left[-\frac{2}{3}{\left(\sqrt{-2x+1}\right)}^3\right]}_{-12}^{-\frac{15}{2}}$

$=$ $-\frac{2}{3}{\left(\sqrt{-2\cdot \left(-\frac{15}{2}\right)+1}\right)}^3+\frac{2}{3}{\left(\sqrt{-2\cdot \left(-12\right)+1}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{15+1}\right)}^3+\frac{2}{3}{\left(\sqrt{24+1}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{16}\right)}^3+\frac{2}{3}{\left(\sqrt{25}\right)}^3$

= $-\frac{2}{3}\cdot 4^3+\frac{2}{3}\cdot 5^3$

= $-\frac{2}{3}\cdot 64+\frac{2}{3}\cdot 125$

= $-\frac{128}{3}+\frac{250}{3}$

= $\frac{122}{3}$

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

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

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

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

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

= $\frac{2}{3}{e}^{\frac{3}{2}\pi }-1·\frac{8}{3}$

= $\frac{2}{3}{e}^{\frac{3}{2}\pi }-\frac{8}{3}$

= ${\left[e^{-2x+3}\right]}_2^4$

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

= $e^{-8+3}-e^{-4+3}$

= $e^{-5}-e^{-1}$