$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

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 +6 = 1.5

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

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

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

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

= $\frac{8}{3}-\frac{12}{3}-\left(\frac{1}{3}-\frac{6}{3}\right)$

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

= $-\frac{4}{3}+\frac{5}{3}$

= $\frac{1}{3}$

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

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

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

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

= $-\frac{3}{2}\cdot 1+\frac{2}{3}{e}^{-12}-\left(-\frac{3}{2}\cdot 1+\frac{2}{3}{e}^{-3}\right)$

= $-\frac{3}{2}+\frac{2}{3}{e}^{-12}-\left(-\frac{3}{2}+\frac{2}{3}{e}^{-3}\right)$

= $\frac{2}{3}{e}^{-12}-\frac{3}{2}-\left(\frac{2}{3}{e}^{-3}-\frac{3}{2}\right)$

= $-\frac{2}{3}{e}^{-3}-1·\left(-\frac{3}{2}\right)+\frac{2}{3}{e}^{-12}-\frac{3}{2}$

= $-\frac{2}{3}{e}^{-3}+\frac{3}{2}+\frac{2}{3}{e}^{-12}-\frac{3}{2}$

= $-\frac{2}{3}{e}^{-3}+\frac{2}{3}{e}^{-12}+\frac{3}{2}-\frac{3}{2}$

= $-\frac{2}{3}{e}^{-3}+\frac{2}{3}{e}^{-12}+0$

= ${\left[\frac{4}{9}{\left(-3x+7\right)}^{\frac{3}{2}}\right]}_{-6}^{-\frac{2}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{-3x+7}\right)}^3\right]}_{-6}^{-\frac{2}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{-3\cdot \left(-\frac{2}{3}\right)+7}\right)}^3-\frac{4}{9}{\left(\sqrt{-3\cdot \left(-6\right)+7}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{2+7}\right)}^3-\frac{4}{9}{\left(\sqrt{18+7}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{9}\right)}^3-\frac{4}{9}{\left(\sqrt{25}\right)}^3$

= $\frac{4}{9}\cdot 3^3-\frac{4}{9}\cdot 5^3$

= $\frac{4}{9}\cdot 27-\frac{4}{9}\cdot 125$

= $12-\frac{500}{9}$

= $\frac{108}{9}-\frac{500}{9}$

= $-\frac{392}{9}$

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

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

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

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

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

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }-1·\frac{8}{3}$

= $-\frac{1}{3}{e}^{\frac{9}{2}\pi }-\frac{8}{3}$

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

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

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

= $\frac{1}{4}\cdot 2^4+27-\left(\frac{1}{4}\cdot {\left(-2\right)}^4+3\right)$

= $\frac{1}{4}\cdot 16+27-\left(\frac{1}{4}\cdot 16+3\right)$

= $4+27-\left(4+3\right)$

= $31-1·7$

= $31-7$

= $24$