$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ 4 = 2 ⋅ 4 = 8.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 7) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 8 +6 -4 -12 = -2

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

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

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

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

= $2-\left(-\frac{9}{2}+\frac{12}{2}\right)$

= $2-1·\frac{3}{2}$

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

= $\frac{4}{2}-\frac{3}{2}$

= $\frac{1}{2}$

= ${\left[\frac{9}{20}{x}^5+\frac{3}{2}{x}^2\right]}_{-3}^1$

$=$ $\frac{9}{20}\cdot 1^5+\frac{3}{2}\cdot 1^2-\left(\frac{9}{20}\cdot {\left(-3\right)}^5+\frac{3}{2}\cdot {\left(-3\right)}^2\right)$

= $\frac{9}{20}\cdot 1+\frac{3}{2}\cdot 1-\left(\frac{9}{20}\cdot \left(-243\right)+\frac{3}{2}\cdot 9\right)$

= $\frac{9}{20}+\frac{3}{2}-\left(-\frac{2187}{20}+\frac{27}{2}\right)$

= $\frac{9}{20}+\frac{30}{20}-\left(-\frac{2187}{20}+\frac{270}{20}\right)$

= $\frac{39}{20}-1·\left(-\frac{1917}{20}\right)$

= $\frac{39}{20}+\frac{1917}{20}$

= $\frac{489}{5}$

= ${\left[\frac{1}{4}{\left(3x-3\right)}^4\right]}_2^5$

$=$ $\frac{1}{4}\cdot {\left(3\cdot 5-3\right)}^4-\frac{1}{4}\cdot {\left(3\cdot 2-3\right)}^4$

= $\frac{1}{4}\cdot {\left(15-3\right)}^4-\frac{1}{4}\cdot {\left(6-3\right)}^4$

= $\frac{1}{4}\cdot {12}^4-\frac{1}{4}\cdot 3^4$

= $\frac{1}{4}\cdot 20736-\frac{1}{4}\cdot 81$

= $5184-\frac{81}{4}$

= $\frac{20736}{4}-\frac{81}{4}$

= $\frac{20655}{4}$

= ${\left[\frac{7}{4}\cdot \cos \left(x\right)-\frac{9}{8}{e}^{2x}\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{7}{4}\cdot \cos \left(\pi \right)-\frac{9}{8}{e}^{2\cdot \pi }-\left(\frac{7}{4}\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{9}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $\frac{7}{4}\cdot \left(-1\right)-\frac{9}{8}{e}^{2\cdot \pi }-\left(\frac{7}{4}\cdot 0-\frac{9}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{7}{4}-\frac{9}{8}{e}^{2\cdot \pi }-\left(0-\frac{9}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{9}{8}{e}^{2\cdot \pi }-\frac{7}{4}+\frac{9}{8}{e}^{\pi }$

= $-\frac{9}{8}{e}^{2\pi }+\frac{9}{8}{e}^{\pi }-\frac{7}{4}$

= ${\left[-\ln \left(\left|$ 3x-5$\right|\right)\right]}_2^3$

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

= $-\ln \left(\left|$ 9-5$\right|\right)+\ln \left(\left|$ 6-5$\right|\right)$

= $-\ln \left(4\right)+\ln \left(\left|$ 6-5$\right|\right)$

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

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

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