$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 6) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 -8 = -14

= ${\left[-\frac{5}{2}{x}^2+4x\right]}_{-3}^1$

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

= $-\frac{5}{2}\cdot 1+4-\left(-\frac{5}{2}\cdot 9-12\right)$

= $-\frac{5}{2}+4-\left(-\frac{45}{2}-12\right)$

= $-\frac{5}{2}+\frac{8}{2}-\left(-\frac{45}{2}-\frac{24}{2}\right)$

= $\frac{3}{2}-1·\left(-\frac{69}{2}\right)$

= $\frac{3}{2}+\frac{69}{2}$

= $36$

= ${\left[\frac{7}{6}{e}^{-3x}-\frac{3}{2}\cdot \cos \left(x\right)\right]}_0^{\pi }$

$=$ $\frac{7}{6}{e}^{-3\cdot \pi }-\frac{3}{2}\cdot \cos \left(\pi \right)-\left(\frac{7}{6}{e}^{-3\cdot \left(0\right)}-\frac{3}{2}\cdot \cos \left(0\right)\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }-\frac{3}{2}\cdot \left(-1\right)-\left(\frac{7}{6}{e}^0-\frac{3}{2}\cdot 1\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{3}{2}-\left(\frac{7}{6}-\frac{3}{2}\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{3}{2}-\left(\frac{7}{6}-\frac{9}{6}\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{3}{2}-1·\left(-\frac{1}{3}\right)$

= $\frac{7}{6}{e}^{-3\cdot \pi }+\frac{3}{2}+\frac{1}{3}$

= $\frac{7}{6}{e}^{-3\pi }+\frac{11}{6}$

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

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

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

= $\frac{1}{3}\cdot 0-\frac{1}{3}\cdot 1^3$

= $0-\frac{1}{3}\cdot 1$

= $0-\frac{1}{3}$

= $0-\frac{1}{3}$

= $-\frac{1}{3}$

= ${\left[-2\cdot \cos \left(x\right)-\frac{5}{4}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-2\cdot \cos \left(\frac{3}{2}\pi \right)-\frac{5}{4}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-2\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{5}{4}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

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

= $0+\frac{5}{4}-\left(0-\frac{5}{4}\right)$

= $0+\frac{5}{4}-\left(0-\frac{5}{4}\right)$

= $\frac{5}{4}+\frac{5}{4}$

= $\frac{5}{2}$

= ${\left[-\frac{3}{2}{e}^{-2x+5}\right]}_2^5$

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

= $-\frac{3}{2}{e}^{-10+5}+\frac{3}{2}{e}^{-4+5}$

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

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