$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-1\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2 ) = -4.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -1.5 -2 -4 = -7.5

= ${\left[\frac{4}{3}{x}^3-5x\right]}_{-3}^{-2}$

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

= $\frac{4}{3}\cdot \left(-8\right)+10-\left(\frac{4}{3}\cdot \left(-27\right)+15\right)$

= $-\frac{32}{3}+10-\left(-36+15\right)$

= $-\frac{32}{3}+\frac{30}{3}-1·\left(-21\right)$

= $-\frac{2}{3}+21$

= $-\frac{2}{3}+\frac{63}{3}$

= $\frac{61}{3}$

= ${\left[-e^x+2x^{-1}\right]}_2^4$

= ${\left[-e^x+\frac{2}{x}\right]}_2^4$

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

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

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

= $-e^4+\frac{1}{2}+e^2-1·1$

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

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

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

= ${\left[-\frac{1}{8}{\left(2x-2\right)}^4\right]}_0^2$

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

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

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

= $-\frac{1}{8}\cdot 16+\frac{1}{8}\cdot 16$

= $-2+2$

= 0

= ${\left[\frac{1}{2}{e}^{-2x}-\frac{9}{4}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{1}{2}{e}^{-2\cdot \pi }-\frac{9}{4}\cdot \cos \left(\pi \right)-\left(\frac{1}{2}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}-\frac{9}{4}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{2}{e}^{-2\cdot \pi }-\frac{9}{4}\cdot \left(-1\right)-\left(\frac{1}{2}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}-\frac{9}{4}\cdot 0\right)$

= $\frac{1}{2}{e}^{-2\cdot \pi }+\frac{9}{4}-\left(\frac{1}{2}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}+0\right)$

= $\frac{1}{2}{e}^{-2\cdot \pi }+\frac{9}{4}-\frac{1}{2}{e}^{-\pi }$

= ${\left[-\frac{2}{3}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{20}{3}}^{\frac{29}{3}}$

= ${\left[-\frac{2}{3}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{20}{3}}^{\frac{29}{3}}$

$=$ $-\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{29}{3}\right)-4}\right)}^3+\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{20}{3}\right)-4}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{29-4}\right)}^3+\frac{2}{3}{\left(\sqrt{20-4}\right)}^3$

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

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

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

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

= $-\frac{122}{3}$