$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -6 -8 -5 = -19

= ${\left[-\frac{5}{3}{x}^3-2x^2\right]}_{-1}^1$

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

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

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

= $-\frac{5}{3}-\frac{6}{3}-\left(\frac{5}{3}-\frac{6}{3}\right)$

= $-\frac{11}{3}-1·\left(-\frac{1}{3}\right)$

= $-\frac{11}{3}+\frac{1}{3}$

= $-\frac{10}{3}$

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

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

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

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

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

= ${\left[\cos (-x-\pi )\right]}_0^{\pi }$

$=$ $\cos (-\pi -\pi )-\cos (-\left(0\right)-\pi )$

= $\cos \left(-2\pi \right)-\cos (-\pi )$

= $1-\left(-1\right)$

= $1+1$

= $2$

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

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

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

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

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

= $-\frac{5}{2}{e}^{3\pi }-\frac{1}{2}+\frac{5}{2}{e}^{\pi }-1·\frac{1}{2}$

= $-\frac{5}{2}{e}^{3\pi }-\frac{1}{2}+\frac{5}{2}{e}^{\pi }-\frac{1}{2}$

= $-\frac{5}{2}{e}^{3\pi }+\frac{5}{2}{e}^{\pi }-\frac{1}{2}-\frac{1}{2}$

= $-\frac{5}{2}{e}^{3\pi }+\frac{5}{2}{e}^{\pi }-1$

= ${\left[-2\ln \left(\left|$ -x+1$\right|\right)\right]}_2^5$

$=$ $-2\ln \left(\left|$ -5+1$\right|\right)+2\ln \left(\left|$ -2+1$\right|\right)$

= $-2\ln \left(\left|$ -5+1$\right|\right)+2\ln \left(\left|$ -2+1$\right|\right)$

= $-2\ln \left(4\right)+2\ln \left(\left|$ -2+1$\right|\right)$

= $-2\ln \left(4\right)+2\ln \left(1\right)$

= $-2\ln \left(4\right)+0$

= $-2\ln \left(4\right)$