$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 1 ) = 3 ⋅ ( - 1 ) = -3.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -1 -3 = -4

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

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

= $\frac{2}{3}\cdot 27+\frac{1}{2}\cdot 9-\left(\frac{2}{3}\cdot \left(-1\right)+\frac{1}{2}\cdot 1\right)$

= $18+\frac{9}{2}-\left(-\frac{2}{3}+\frac{1}{2}\right)$

= $\frac{36}{2}+\frac{9}{2}-\left(-\frac{4}{6}+\frac{3}{6}\right)$

= $\frac{45}{2}-1·\left(-\frac{1}{6}\right)$

= $\frac{45}{2}+\frac{1}{6}$

= $\frac{135}{6}+\frac{1}{6}$

= $\frac{68}{3}$

= ${\left[-\frac{16}{5}{x}^{\frac{5}{4}}+e^{-2x}\right]}_4^9$

= ${\left[-\frac{16}{5}{\left(\sqrt[4]{x}\right)}^5+e^{-2x}\right]}_4^9$

$=$ $-\frac{16}{5}{\left(\sqrt[4]{9}\right)}^5+e^{-2\cdot 9}-\left(-\frac{16}{5}{\left(\sqrt[4]{4}\right)}^5+e^{-2\cdot 4}\right)$

= $-\frac{16}{5}\cdot 1^5+e^{-18}-\left(-\frac{16}{5}\cdot 1^5+e^{-8}\right)$

= $-\frac{16}{5}\cdot 1+e^{-18}-\left(-\frac{16}{5}\cdot 1+e^{-8}\right)$

= $-\frac{16}{5}+e^{-18}-\left(-\frac{16}{5}+e^{-8}\right)$

= $e^{-18}-\frac{16}{5}-\left(e^{-8}-\frac{16}{5}\right)$

= $-e^{-8}-1·\left(-\frac{16}{5}\right)+e^{-18}-\frac{16}{5}$

= $-e^{-8}+\frac{16}{5}+e^{-18}-\frac{16}{5}$

= $-e^{-8}+e^{-18}+\frac{16}{5}-\frac{16}{5}$

= $-e^{-8}+e^{-18}+0$

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

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

= $-\cos \left(-4\pi \right)+\cos \left(-\frac{5}{2}\pi \right)$

= $-1+0$

= $-1$

= ${\left[\frac{1}{5}{x}^5-x^{\frac{3}{2}}\right]}_1^4$

= ${\left[\frac{1}{5}{x}^5-{\left(\sqrt{x}\right)}^3\right]}_1^4$

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

= $\frac{1}{5}\cdot 1024-2^3-\left(\frac{1}{5}\cdot 1-1^3\right)$

= $\frac{1024}{5}-8-\left(\frac{1}{5}-1\right)$

= $\frac{1024}{5}-8-1·\frac{1}{5}-1·\left(-1\right)$

= $\frac{1024}{5}-8-\frac{1}{5}+1$

= $\frac{1024}{5}-\frac{40}{5}-\frac{1}{5}+\frac{5}{5}$

= $\frac{988}{5}$

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

$=$ $\frac{1}{3}{e}^{-3\cdot 3+6}-\frac{1}{3}{e}^{-3\cdot 0+6}$

= $\frac{1}{3}{e}^{-9+6}-\frac{1}{3}{e}^{0+6}$

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