$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 3 = 3 ⋅ 3 = 9.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{3\; +\; <mn>1</mn>}}{2}$ = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4.5 +9 +4 = 17.5

= ${\left[\frac{1}{2}{x}^2+4x\right]}_{-1}^0$

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

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

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

= $0-\left(\frac{1}{2}-\frac{8}{2}\right)$

= $-1·\left(-\frac{7}{2}\right)$

= $\frac{7}{2}$

= ${\left[-\frac{1}{6}{e}^{-2x}-\frac{2}{3}{x}^{\frac{3}{2}}\right]}_0^1$

= ${\left[-\frac{1}{6}{e}^{-2x}-\frac{2}{3}{\left(\sqrt{x}\right)}^3\right]}_0^1$

$=$ $-\frac{1}{6}{e}^{-2\cdot 1}-\frac{2}{3}{\left(\sqrt{1}\right)}^3-\left(-\frac{1}{6}{e}^{-2\cdot 0}-\frac{2}{3}{\left(\sqrt{0}\right)}^3\right)$

= $-\frac{1}{6}{e}^{-2}-\frac{2}{3}\cdot 1^3-\left(-\frac{1}{6}{e}^0-\frac{2}{3}\cdot 0^3\right)$

= $-\frac{1}{6}{e}^{-2}-\frac{2}{3}\cdot 1-\left(-\frac{1}{6}-\frac{2}{3}\cdot 0\right)$

= $-\frac{1}{6}{e}^{-2}-\frac{2}{3}-\left(-\frac{1}{6}+0\right)$

= $-\frac{1}{6}{e}^{-2}-\frac{2}{3}-\left(-\frac{1}{6}+0\right)$

= $-\frac{1}{6}{e}^{-2}-\frac{2}{3}+\frac{1}{6}$

= $-\frac{1}{6}{e}^{-2}-\frac{1}{2}$

= ${\left[\frac{2}{3}{e}^{-3x+7}\right]}_0^2$

$=$ $\frac{2}{3}{e}^{-3\cdot 2+7}-\frac{2}{3}{e}^{-3\cdot 0+7}$

= $\frac{2}{3}{e}^{-6+7}-\frac{2}{3}{e}^{0+7}$

= $\frac{2}{3}e-\frac{2}{3}{e}^7$

= $\frac{2}{3}e-\frac{2}{3}{e}^7$

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

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

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

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

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

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

= $-1·2-1·\left(-\frac{4}{3{\pi }^2}\right)-\frac{1}{3{\pi }^2}$

= $-2+\frac{4}{3{\pi }^2}-\frac{1}{3{\pi }^2}$

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

= ${\left[-\frac{1}{3}{e}^{-3x+7}\right]}_1^2$

$=$ $-\frac{1}{3}{e}^{-3\cdot 2+7}+\frac{1}{3}{e}^{-3\cdot 1+7}$

= $-\frac{1}{3}{e}^{-6+7}+\frac{1}{3}{e}^{-3+7}$

= $-\frac{1}{3}e+\frac{1}{3}{e}^4$

= $-\frac{1}{3}e+\frac{1}{3}{e}^4$