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

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

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

= $-2\cdot 9-6-\left(-2\cdot 1+2\right)$

= $-18-6-\left(-2+2\right)$

= $-24-1·0$

= $-24+0$

= $-24$

= ${\left[\frac{5}{2}{x}^{-2}+\frac{1}{2}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

= ${\left[\frac{5}{2x^2}+\frac{1}{2}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

= $\frac{5}{2{\pi }^2}-\left(\frac{1}{2}+\frac{10}{{\pi }^2}\right)$

= $-1·\frac{1}{2}-1·\frac{10}{{\pi }^2}+\frac{5}{2{\pi }^2}$

= $-\frac{1}{2}-\frac{10}{{\pi }^2}+\frac{5}{2{\pi }^2}$

= $-\frac{1}{2}-\frac{15}{2{\pi }^2}$

= ${\left[-\frac{2}{3}{\left(-x+2\right)}^{\frac{3}{2}}\right]}_{-23}^{-7}$

= ${\left[-\frac{2}{3}{\left(\sqrt{-x+2}\right)}^3\right]}_{-23}^{-7}$

$=$ $-\frac{2}{3}{\left(\sqrt{-\left(-7\right)+2}\right)}^3+\frac{2}{3}{\left(\sqrt{-\left(-23\right)+2}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{7+2}\right)}^3+\frac{2}{3}{\left(\sqrt{23+2}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{9}\right)}^3+\frac{2}{3}{\left(\sqrt{25}\right)}^3$

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

= $-\frac{2}{3}\cdot 27+\frac{2}{3}\cdot 125$

= $-18+\frac{250}{3}$

= $-\frac{54}{3}+\frac{250}{3}$

= $\frac{196}{3}$

= ${\left[\frac{7}{3}{x}^3-\frac{3}{2}{x}^{\frac{4}{3}}\right]}_1^9$

= ${\left[\frac{7}{3}{x}^3-\frac{3}{2}{\left(\sqrt[3]{x}\right)}^4\right]}_1^9$

$=$ $\frac{7}{3}\cdot 9^3-\frac{3}{2}{\left(\sqrt[3]{9}\right)}^4-\left(\frac{7}{3}\cdot 1^3-\frac{3}{2}{\left(\sqrt[3]{1}\right)}^4\right)$

= $\frac{7}{3}\cdot 729-\frac{3}{2}\cdot 1^4-\left(\frac{7}{3}\cdot 1-\frac{3}{2}\cdot 1^4\right)$

= $1701-\frac{3}{2}\cdot 1-\left(\frac{7}{3}-\frac{3}{2}\cdot 1\right)$

= $1701-\frac{3}{2}-\left(\frac{7}{3}-\frac{3}{2}\right)$

= $\frac{3402}{2}-\frac{3}{2}-\left(\frac{14}{6}-\frac{9}{6}\right)$

= $\frac{3399}{2}-1·\frac{5}{6}$

= $\frac{3399}{2}-\frac{5}{6}$

= $\frac{10197}{6}-\frac{5}{6}$

= $\frac{3399}{2}-\frac{5}{6}$

= $\frac{10197}{6}-\frac{5}{6}$

= $\frac{5096}{3}$

= ${\left[\frac{1}{9}{\left(-3x+4\right)}^3\right]}_2^5$

$=$ $\frac{1}{9}\cdot {\left(-3\cdot 5+4\right)}^3-\frac{1}{9}\cdot {\left(-3\cdot 2+4\right)}^3$

= $\frac{1}{9}\cdot {\left(-15+4\right)}^3-\frac{1}{9}\cdot {\left(-6+4\right)}^3$

= $\frac{1}{9}\cdot {\left(-11\right)}^3-\frac{1}{9}\cdot {\left(-2\right)}^3$

= $\frac{1}{9}\cdot \left(-1331\right)-\frac{1}{9}\cdot \left(-8\right)$

= $-\frac{1331}{9}+\frac{8}{9}$

= $-147$