$$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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:

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:

$$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = 3 = 3

= ${\left[2x^2+4x\right]}_{-1}^0$

$=$ $2\cdot 0^2+4\cdot 0-\left(2\cdot {\left(-1\right)}^2+4\cdot \left(-1\right)\right)$

= $2\cdot 0+0-\left(2\cdot 1-4\right)$

= $0+0-\left(2-4\right)$

= $0-1·\left(-2\right)$

= $2$

= ${\left[9\cdot \cos \left(x\right)+\frac{1}{3}{e}^{-3x}\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $9\cdot \cos \left(\pi \right)+\frac{1}{3}{e}^{-3\cdot \pi }-\left(9\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{1}{3}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $9\cdot \left(-1\right)+\frac{1}{3}{e}^{-3\cdot \pi }-\left(9\cdot 0+\frac{1}{3}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-9+\frac{1}{3}{e}^{-3\cdot \pi }-\left(0+\frac{1}{3}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

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

= ${\left[\frac{2}{3}{\left(x-3\right)}^{\frac{3}{2}}\right]}_{19}^{28}$

= ${\left[\frac{2}{3}{\left(\sqrt{x-3}\right)}^3\right]}_{19}^{28}$

$=$ $\frac{2}{3}{\left(\sqrt{28-3}\right)}^3-\frac{2}{3}{\left(\sqrt{19-3}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{16}\right)}^3$

= $\frac{2}{3}\cdot 5^3-\frac{2}{3}\cdot 4^3$

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 64$

= $\frac{250}{3}-\frac{128}{3}$

= $\frac{122}{3}$

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

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

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

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

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

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

= $-\frac{15}{20}-\frac{48}{20}-\left(-\frac{15}{5}-\frac{12}{5}\right)$

= $-\frac{63}{20}-1·\left(-\frac{27}{5}\right)$

= $-\frac{63}{20}+\frac{27}{5}$

= $-\frac{63}{20}+\frac{108}{20}$

= $-\frac{63}{20}+\frac{27}{5}$

= $-\frac{63}{20}+\frac{108}{20}$

= $\frac{9}{4}$

= ${\left[-\frac{1}{8}{\left(2x-3\right)}^4\right]}_1^3$

$=$ $-\frac{1}{8}\cdot {\left(2\cdot 3-3\right)}^4+\frac{1}{8}\cdot {\left(2\cdot 1-3\right)}^4$

= $-\frac{1}{8}\cdot {\left(6-3\right)}^4+\frac{1}{8}\cdot {\left(2-3\right)}^4$

= $-\frac{1}{8}\cdot 3^4+\frac{1}{8}\cdot {\left(-1\right)}^4$

= $-\frac{1}{8}\cdot 81+\frac{1}{8}\cdot 1$

= $-\frac{81}{8}+\frac{1}{8}$

= $-10$