$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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

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

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

= $-1+4-\left(-9-12\right)$

= $-1+4-1·\left(-9\right)-1·\left(-12\right)$

= $-1+4+9+12$

= $24$

= ${\left[-\frac{8}{9}{x}^3+\frac{2}{3}{e}^{3x}\right]}_1^5$

$=$ $-\frac{8}{9}\cdot 5^3+\frac{2}{3}{e}^{3\cdot 5}-\left(-\frac{8}{9}\cdot 1^3+\frac{2}{3}{e}^{3\cdot 1}\right)$

= $-\frac{8}{9}\cdot 125+\frac{2}{3}{e}^{15}-\left(-\frac{8}{9}\cdot 1+\frac{2}{3}{e}^3\right)$

= $-\frac{1000}{9}+\frac{2}{3}{e}^{15}-\left(-\frac{8}{9}+\frac{2}{3}{e}^3\right)$

= $\frac{2}{3}{e}^{15}-\frac{1000}{9}-\left(\frac{2}{3}{e}^3-\frac{8}{9}\right)$

= $\frac{2}{3}{e}^{15}-\frac{1000}{9}-\frac{2}{3}{e}^3-1·\left(-\frac{8}{9}\right)$

= $\frac{2}{3}{e}^{15}-\frac{1000}{9}-\frac{2}{3}{e}^3+\frac{8}{9}$

= $\frac{2}{3}{e}^{15}-\frac{2}{3}{e}^3-\frac{1000}{9}+\frac{8}{9}$

= $\frac{2}{3}{e}^{15}-\frac{2}{3}{e}^3-\frac{992}{9}$

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

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

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

= $-\frac{3}{8}\cdot 2^4+2-\left(-\frac{3}{8}\cdot 0^4+\frac{1}{2}\right)$

= $-\frac{3}{8}\cdot 16+2-\left(-\frac{3}{8}\cdot 0+\frac{1}{2}\right)$

= $-6+2-\left(0+\frac{1}{2}\right)$

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

= $-4-\frac{1}{2}$

= $-\frac{8}{2}-\frac{1}{2}$

= $-4-\frac{1}{2}$

= $-\frac{8}{2}-\frac{1}{2}$

= $-\frac{9}{2}$

= ${\left[-2\cdot \sin \left(x\right)+\frac{7}{4}\cdot \cos \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $-2\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{7}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(-2\cdot \sin \left(0\right)+\frac{7}{4}\cdot \cos \left(0\right)\right)$

= $-2\cdot \left(-1\right)+\frac{7}{4}\cdot 0-\left(-2\cdot 0+\frac{7}{4}\cdot 1\right)$

= $2+0-\left(0+\frac{7}{4}\right)$

= $2-\left(0+\frac{7}{4}\right)$

= $2-\frac{7}{4}$

= $\frac{8}{4}-\frac{7}{4}$

= $\frac{1}{4}$

= ${\left[\frac{1}{3}\cdot \cos \left(-3x+\frac{3}{2}\pi \right)\right]}_0^{\frac{3}{2}\pi }$

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

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

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

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

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

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