$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ 3 = 3 ⋅ 3 = 9.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4.5 +9 = 13.5

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

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

= $\frac{1}{2}\cdot 25+5-\left(\frac{1}{2}\cdot 1+1\right)$

= $\frac{25}{2}+5-\left(\frac{1}{2}+1\right)$

= $\frac{25}{2}+\frac{10}{2}-\left(\frac{1}{2}+\frac{2}{2}\right)$

= $\frac{35}{2}-1·\frac{3}{2}$

= $\frac{35}{2}-\frac{3}{2}$

= $16$

= ${\left[-2\cdot \cos \left(x\right)+8\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $-2\cdot \cos \left(\frac{1}{2}\pi \right)+8\cdot \sin \left(\frac{1}{2}\pi \right)-\left(-2\cdot \cos \left(0\right)+8\cdot \sin \left(0\right)\right)$

= $-2\cdot 0+8\cdot 1-\left(-2\cdot 1+8\cdot 0\right)$

= $0+8-\left(-2+0\right)$

= $8+2$

= $10$

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

$=$ $e^{-2\cdot 3+4}-e^{-2\cdot 1+4}$

= $e^{-6+4}-e^{-2+4}$

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

= ${\left[\frac{5}{3}{x}^{-3}+\frac{1}{3}{e}^{2x}\right]}_2^5$

= ${\left[\frac{5}{3x^3}+\frac{1}{3}{e}^{2x}\right]}_2^5$

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

= $\frac{5}{3}\cdot \left(\frac{1}{125}\right)+\frac{1}{3}{e}^{10}-\left(\frac{5}{3}\cdot \left(\frac{1}{8}\right)+\frac{1}{3}{e}^4\right)$

= $\frac{1}{75}+\frac{1}{3}{e}^{10}-\left(\frac{5}{24}+\frac{1}{3}{e}^4\right)$

= $\frac{1}{3}{e}^{10}+\frac{1}{75}-\left(\frac{1}{3}{e}^4+\frac{5}{24}\right)$

= $\frac{1}{3}{e}^{10}+\frac{1}{75}-\frac{1}{3}{e}^4-1·\frac{5}{24}$

= $\frac{1}{3}{e}^{10}+\frac{1}{75}-\frac{1}{3}{e}^4-\frac{5}{24}$

= $\frac{1}{3}{e}^{10}-\frac{1}{3}{e}^4+\frac{1}{75}-\frac{5}{24}$

= $\frac{1}{3}{e}^{10}-\frac{1}{3}{e}^4-\frac{39}{200}$

= ${\left[\frac{1}{8}{\left(-2x+1\right)}^4+x^2\right]}_2^5$

$=$ $\frac{1}{8}\cdot {\left(-2\cdot 5+1\right)}^4+5^2-\left(\frac{1}{8}\cdot {\left(-2\cdot 2+1\right)}^4+2^2\right)$

= $\frac{1}{8}\cdot {\left(-10+1\right)}^4+25-\left(\frac{1}{8}\cdot {\left(-4+1\right)}^4+4\right)$

= $\frac{1}{8}\cdot {\left(-9\right)}^4+25-\left(\frac{1}{8}\cdot {\left(-3\right)}^4+4\right)$

= $\frac{1}{8}\cdot 6561+25-\left(\frac{1}{8}\cdot 81+4\right)$

= $\frac{6561}{8}+25-\left(\frac{81}{8}+4\right)$

= $\frac{6561}{8}+\frac{200}{8}-\left(\frac{81}{8}+\frac{32}{8}\right)$

= $\frac{6761}{8}-1·\frac{113}{8}$

= $\frac{6761}{8}-\frac{113}{8}$

= $831$