$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

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.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -4 -3 +3 = -4

= ${\left[-\frac{1}{2}{x}^2+4x\right]}_{-3}^0$

$=$ $-\frac{1}{2}\cdot 0^2+4\cdot 0-\left(-\frac{1}{2}\cdot {\left(-3\right)}^2+4\cdot \left(-3\right)\right)$

= $-\frac{1}{2}\cdot 0+0-\left(-\frac{1}{2}\cdot 9-12\right)$

= $0+0-\left(-\frac{9}{2}-12\right)$

= $0-\left(-\frac{9}{2}-\frac{24}{2}\right)$

= $-1·\left(-\frac{33}{2}\right)$

= $\frac{33}{2}$

= ${\left[-\frac{1}{5}{x}^{\frac{5}{4}}-\frac{3}{8}{e}^{-2x}\right]}_0^1$

= ${\left[-\frac{1}{5}{\left(\sqrt[4]{x}\right)}^5-\frac{3}{8}{e}^{-2x}\right]}_0^1$

$=$ $-\frac{1}{5}{\left(\sqrt[4]{1}\right)}^5-\frac{3}{8}{e}^{-2\cdot 1}-\left(-\frac{1}{5}{\left(\sqrt[4]{0}\right)}^5-\frac{3}{8}{e}^{-2\cdot 0}\right)$

= $-\frac{1}{5}\cdot 1^5-\frac{3}{8}{e}^{-2}-\left(-\frac{1}{5}\cdot 0^5-\frac{3}{8}{e}^0\right)$

= $-\frac{1}{5}\cdot 1-\frac{3}{8}{e}^{-2}-\left(-\frac{1}{5}\cdot 0-\frac{3}{8}\right)$

= $-\frac{1}{5}-\frac{3}{8}{e}^{-2}-\left(0-\frac{3}{8}\right)$

= $-\frac{3}{8}{e}^{-2}-\frac{1}{5}-\left(0-\frac{3}{8}\right)$

= $-\frac{3}{8}{e}^{-2}-\frac{1}{5}+\frac{3}{8}$

= $-\frac{3}{8}{e}^{-2}+\frac{7}{40}$

= ${\left[\frac{1}{12}{\left(-3x+5\right)}^4+\frac{5}{2}{x}^2\right]}_0^2$

$=$ $\frac{1}{12}\cdot {\left(-3\cdot 2+5\right)}^4+\frac{5}{2}\cdot 2^2-\left(\frac{1}{12}\cdot {\left(-3\cdot 0+5\right)}^4+\frac{5}{2}\cdot 0^2\right)$

= $\frac{1}{12}\cdot {\left(-6+5\right)}^4+\frac{5}{2}\cdot 4-\left(\frac{1}{12}\cdot {\left(0+5\right)}^4+\frac{5}{2}\cdot 0\right)$

= $\frac{1}{12}\cdot {\left(-1\right)}^4+10-\left(\frac{1}{12}\cdot 5^4+0\right)$

= $\frac{1}{12}\cdot 1+10-\left(\frac{1}{12}\cdot 625+0\right)$

= $\frac{1}{12}+10-\left(\frac{625}{12}+0\right)$

= $\frac{1}{12}+\frac{120}{12}-\left(\frac{625}{12}+0\right)$

= $\frac{121}{12}-\frac{625}{12}$

= $-42$

= ${\left[4\cdot \sin \left(x\right)-\frac{7}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $4\cdot \sin \left(\pi \right)-\frac{7}{3}\cdot \cos \left(\pi \right)-\left(4\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{7}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $4\cdot 0-\frac{7}{3}\cdot \left(-1\right)-\left(4\cdot 1-\frac{7}{3}\cdot 0\right)$

= $0+\frac{7}{3}-\left(4+0\right)$

= $0+\frac{7}{3}-4$

= $\frac{7}{3}-4$

= $\frac{7}{3}-\frac{12}{3}$

= $-\frac{5}{3}$

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

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

= $2\cdot \sin \left(-2\pi \right)-2\cdot \sin (-\pi )$

= $2\cdot 0-2\cdot 0$

= $0+0$

= 0