$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

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

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 +4 = 7

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

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

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

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

= $-\frac{1}{2}-\frac{10}{2}+0$

= $-\frac{11}{2}$

= ${\left[\frac{3}{4}{x}^{-1}-\frac{12}{5}{x}^{\frac{5}{4}}\right]}_1^{16}$

= ${\left[\frac{3}{4x}-\frac{12}{5}{\left(\sqrt[4]{x}\right)}^5\right]}_1^{16}$

$=$ $\frac{3}{4\cdot 16}-\frac{12}{5}{\left(\sqrt[4]{16}\right)}^5-\left(\frac{3}{4}-\frac{12}{5}{\left(\sqrt[4]{1}\right)}^5\right)$

= $\frac{3}{4}\cdot \left(\frac{1}{16}\right)-\frac{12}{5}\cdot 2^5-\left(\frac{3}{4}\cdot 1-\frac{12}{5}\cdot 1^5\right)$

= $\frac{3}{64}-\frac{12}{5}\cdot 32-\left(\frac{3}{4}-\frac{12}{5}\cdot 1\right)$

= $\frac{3}{64}-\frac{384}{5}-\left(\frac{3}{4}-\frac{12}{5}\right)$

= $\frac{15}{320}-\frac{24576}{320}-\left(\frac{15}{20}-\frac{48}{20}\right)$

= $-\frac{24561}{320}-1·\left(-\frac{33}{20}\right)$

= $-\frac{24561}{320}+\frac{33}{20}$

= $-\frac{24561}{320}+\frac{528}{320}$

= $-\frac{24561}{320}+\frac{33}{20}$

= $-\frac{24561}{320}+\frac{528}{320}$

= $-\frac{24033}{320}$

= ${\left[{\left(2x-2\right)}^{\frac{3}{2}}\right]}_{\frac{11}{2}}^9$

= ${\left[{\left(\sqrt{2x-2}\right)}^3\right]}_{\frac{11}{2}}^9$

$=$ ${\left(\sqrt{2\cdot 9-2}\right)}^3-{\left(\sqrt{2\cdot \left(\frac{11}{2}\right)-2}\right)}^3$

= ${\left(\sqrt{18-2}\right)}^3-{\left(\sqrt{11-2}\right)}^3$

= ${\left(\sqrt{16}\right)}^3-{\left(\sqrt{9}\right)}^3$

= $4^3-3^3$

= $64-27$

= $37$

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

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

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

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

= $-4-\frac{9}{2}\cdot {\pi }^3-\left(4-\frac{1}{6}\cdot {\pi }^3\right)$

= $-4-\frac{9}{2}\cdot {\pi }^3-1·4-1·\left(-\frac{1}{6}\cdot {\pi }^3\right)$

= $-4-\frac{9}{2}\cdot {\pi }^3-4+\frac{1}{6}\cdot {\pi }^3$

= $-4-4-\frac{9}{2}\cdot {\pi }^3+\frac{1}{6}\cdot {\pi }^3$

= $-8-\frac{13}{3}\cdot {\pi }^3$

= ${\left[-2e^{x-2}\right]}_1^3$

$=$ $-2e^{3-2}+2e^{1-2}$

= $-2e+2e^{-1}$

= $-2e+2e^{-1}$