$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 +1.5 +2 = -0.5

= ${\left[-x^3+3x\right]}_0^1$

$=$ $-1^3+3\cdot 1-\left(-0^3+3\cdot 0\right)$

= $-1+3-\left(-0+0\right)$

= $-1+3$

= $2$

= ${\left[-\frac{5}{3}\cdot \sin \left(x\right)-\frac{3}{8}{e}^{-2x}\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{3}\cdot \sin \left(\pi \right)-\frac{3}{8}{e}^{-2\cdot \pi }-\left(-\frac{5}{3}\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{3}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{5}{3}\cdot 0-\frac{3}{8}{e}^{-2\cdot \pi }-\left(-\frac{5}{3}\cdot 1-\frac{3}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $0-\frac{3}{8}{e}^{-2\cdot \pi }-\left(-\frac{5}{3}-\frac{3}{8}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{3}{8}{e}^{-2\pi }-\left(-\frac{3}{8}{e}^{-\pi }-\frac{5}{3}\right)$

= $-\frac{3}{8}{e}^{-2\pi }+\frac{3}{8}{e}^{-\pi }-1·\left(-\frac{5}{3}\right)$

= $-\frac{3}{8}{e}^{-2\pi }+\frac{3}{8}{e}^{-\pi }+\frac{5}{3}$

= $\frac{3}{8}{e}^{-\pi }-\frac{3}{8}{e}^{-2\pi }+\frac{5}{3}$

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

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

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

= $\frac{1}{2}\cdot {\left(-7\right)}^3+8-\left(\frac{1}{2}\cdot {\left(-1\right)}^3+2\right)$

= $\frac{1}{2}\cdot \left(-343\right)+8-\left(\frac{1}{2}\cdot \left(-1\right)+2\right)$

= $-\frac{343}{2}+8-\left(-\frac{1}{2}+2\right)$

= $-\frac{343}{2}+\frac{16}{2}-\left(-\frac{1}{2}+\frac{4}{2}\right)$

= $-\frac{327}{2}-1·\frac{3}{2}$

= $-\frac{327}{2}-\frac{3}{2}$

= $-165$

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

$=$ $\frac{3}{2}\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{3}{2}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\frac{3}{2}\cdot \sin \left(0\right)+\frac{3}{2}\cdot \cos \left(0\right)\right)$

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

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

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

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

= $-3$

= ${\left[-\ln \left(\left|$ -2x+3$\right|\right)\right]}_3^5$

$=$ $-\ln \left(\left|$ -2\cdot 5+3$\right|\right)+\ln \left(\left|$ -2\cdot 3+3$\right|\right)$

= $-\ln \left(\left|$ -10+3$\right|\right)+\ln \left(\left|$ -6+3$\right|\right)$

= $-\ln \left(7\right)+\ln \left(\left|$ -6+3$\right|\right)$

= $-\ln \left(7\right)+\ln \left(3\right)$