$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

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

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

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4.5 +3 +9 = 7.5

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

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

= $\frac{5}{2}\cdot 1-5-\left(\frac{5}{2}\cdot 9+15\right)$

= $\frac{5}{2}-5-\left(\frac{45}{2}+15\right)$

= $\frac{5}{2}-\frac{10}{2}-\left(\frac{45}{2}+\frac{30}{2}\right)$

= $-\frac{5}{2}-1·\frac{75}{2}$

= $-\frac{5}{2}-\frac{75}{2}$

= $-40$

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

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

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

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

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

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

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

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

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

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

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

= $0+0$

= 0

= ${\left[7\cdot \sin \left(x\right)+5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $7\cdot \sin \left(\pi \right)+5\cdot \cos \left(\pi \right)-\left(7\cdot \sin \left(\frac{1}{2}\pi \right)+5\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $7\cdot 0+5\cdot \left(-1\right)-\left(7\cdot 1+5\cdot 0\right)$

= $0-5-\left(7+0\right)$

= $-5-7$

= $-12$

= ${\left[-\frac{1}{4}{\left(-3x+4\right)}^4+5x\right]}_2^3$

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

= $-\frac{1}{4}\cdot {\left(-9+4\right)}^4+15-\left(-\frac{1}{4}\cdot {\left(-6+4\right)}^4+10\right)$

= $-\frac{1}{4}\cdot {\left(-5\right)}^4+15-\left(-\frac{1}{4}\cdot {\left(-2\right)}^4+10\right)$

= $-\frac{1}{4}\cdot 625+15-\left(-\frac{1}{4}\cdot 16+10\right)$

= $-\frac{625}{4}+15-\left(-4+10\right)$

= $-\frac{625}{4}+\frac{60}{4}-1·6$

= $-\frac{565}{4}-6$

= $-\frac{565}{4}-\frac{24}{4}$

= $-\frac{565}{4}-6$

= $-\frac{565}{4}-\frac{24}{4}$

= $-\frac{589}{4}$