$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 3 = 3 ⋅ 3 = 9.

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.

Somit gilt:

$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = 9 +3 = 12

= ${\left[-\frac{3}{2}{x}^2+x\right]}_1^4$

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

= $-\frac{3}{2}\cdot 16+4-\left(-\frac{3}{2}\cdot 1+1\right)$

= $-24+4-\left(-\frac{3}{2}+1\right)$

= $-20-\left(-\frac{3}{2}+\frac{2}{2}\right)$

= $-20-1·\left(-\frac{1}{2}\right)$

= $-20+\frac{1}{2}$

= $-\frac{40}{2}+\frac{1}{2}$

= $-\frac{39}{2}$

= ${\left[\frac{1}{2}\cdot \cos \left(x\right)+\frac{5}{12}{x}^6\right]}_0^{\pi }$

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

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

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

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

= $-\frac{1}{2}+\frac{5}{12}\cdot {\pi }^6-\frac{1}{2}$

= $-\frac{1}{2}-\frac{1}{2}+\frac{5}{12}\cdot {\pi }^6$

= $-1+\frac{5}{12}\cdot {\pi }^6$

= ${\left[{\left(x-2\right)}^3\right]}_0^2$

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

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

= $0-\left(-8\right)$

= $0+8$

= $8$

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

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

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

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

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

= $\frac{2}{3}{e}^{-\frac{9}{2}\pi }-2-\frac{2}{3}{e}^{-\frac{3}{2}\pi }-1·2$

= $\frac{2}{3}{e}^{-\frac{9}{2}\pi }-2-\frac{2}{3}{e}^{-\frac{3}{2}\pi }-2$

= $-\frac{2}{3}{e}^{-\frac{3}{2}\pi }+\frac{2}{3}{e}^{-\frac{9}{2}\pi }-2-2$

= $-\frac{2}{3}{e}^{-\frac{3}{2}\pi }+\frac{2}{3}{e}^{-\frac{9}{2}\pi }-4$

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

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

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

= $-0+0$

= 0