$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

I₄ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 6) ⋅ $\frac{\mathrm{-1\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 1.5 ) = -4.5.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -1 -2 -4.5 = -7.5

= ${\left[\frac{5}{3}{x}^3+x^2\right]}_0^2$

$=$ $\frac{5}{3}\cdot 2^3+2^2-\left(\frac{5}{3}\cdot 0^3+0^2\right)$

= $\frac{5}{3}\cdot 8+4-\left(\frac{5}{3}\cdot 0+0\right)$

= $\frac{40}{3}+4-\left(0+0\right)$

= $\frac{40}{3}+\frac{12}{3}+0$

= $\frac{52}{3}$

= ${\left[-\frac{7}{3}\cdot \cos \left(x\right)+3\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{7}{3}\cdot \cos \left(\pi \right)+3\cdot \sin \left(\pi \right)-\left(-\frac{7}{3}\cdot \cos \left(\frac{1}{2}\pi \right)+3\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{7}{3}\cdot \left(-1\right)+3\cdot 0-\left(-\frac{7}{3}\cdot 0+3\cdot 1\right)$

= $\frac{7}{3}+0-\left(0+3\right)$

= $\frac{7}{3}+0-3$

= $\frac{7}{3}-3$

= $\frac{7}{3}-\frac{9}{3}$

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

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

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

= $-\frac{1}{2}{e}^{-6+1}+\frac{1}{2}{e}^{-2+1}$

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

= ${\left[-\frac{3}{2}\cdot \cos \left(x\right)-\frac{5}{3}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{3}{2}\cdot \cos \left(\frac{3}{2}\pi \right)-\frac{5}{3}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{3}{2}\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{5}{3}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{3}{2}\cdot 0-\frac{5}{3}\cdot \left(-1\right)-\left(-\frac{3}{2}\cdot 0-\frac{5}{3}\cdot 1\right)$

= $0+\frac{5}{3}-\left(0-\frac{5}{3}\right)$

= $0+\frac{5}{3}-\left(0-\frac{5}{3}\right)$

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

= $\frac{10}{3}$

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

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

= $-\frac{3}{8}\cdot {\left(6-3\right)}^4-\frac{1}{2}\cdot 9-\left(-\frac{3}{8}\cdot {\left(2-3\right)}^4-\frac{1}{2}\cdot 1\right)$

= $-\frac{3}{8}\cdot 3^4-\frac{9}{2}-\left(-\frac{3}{8}\cdot {\left(-1\right)}^4-\frac{1}{2}\right)$

= $-\frac{3}{8}\cdot 81-\frac{9}{2}-\left(-\frac{3}{8}\cdot 1-\frac{1}{2}\right)$

= $-\frac{243}{8}-\frac{9}{2}-\left(-\frac{3}{8}-\frac{1}{2}\right)$

= $-\frac{243}{8}-\frac{36}{8}-\left(-\frac{3}{8}-\frac{4}{8}\right)$

= $-\frac{279}{8}-1·\left(-\frac{7}{8}\right)$

= $-\frac{279}{8}+\frac{7}{8}$

= $-34$