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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 4 = 2 ⋅ 4 = 8.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -2 +6 +8 = 12

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

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

= $\frac{4}{3}\cdot \left(-1\right)+1-\left(\frac{4}{3}\cdot \left(-27\right)+9\right)$

= $-\frac{4}{3}+1-\left(-36+9\right)$

= $-\frac{4}{3}+\frac{3}{3}-1·\left(-27\right)$

= $-\frac{1}{3}+27$

= $-\frac{1}{3}+\frac{81}{3}$

= $\frac{80}{3}$

= ${\left[-x^{-1}-\frac{5}{24}{x}^6\right]}_1^5$

= ${\left[-\frac{1}{x}-\frac{5}{24}{x}^6\right]}_1^5$

$=$ $-\frac{1}{5}-\frac{5}{24}\cdot 5^6-\left(-\frac{1}{1}-\frac{5}{24}\cdot 1^6\right)$

= $-\left(\frac{1}{5}\right)-\frac{5}{24}\cdot 15625-\left(-1-\frac{5}{24}\cdot 1\right)$

= $-\frac{1}{5}-\frac{78125}{24}-\left(-1-\frac{5}{24}\right)$

= $-\frac{24}{120}-\frac{390625}{120}-\left(-\frac{24}{24}-\frac{5}{24}\right)$

= $-\frac{390649}{120}-1·\left(-\frac{29}{24}\right)$

= $-\frac{390649}{120}+\frac{29}{24}$

= $-\frac{390649}{120}+\frac{145}{120}$

= $-\frac{16271}{5}$

= ${\left[-\frac{1}{2}{e}^{-2x+5}\right]}_1^3$

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

= $-\frac{1}{2}{e}^{-6+5}+\frac{1}{2}{e}^{-2+5}$

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

= ${\left[e^{3x}-\frac{1}{2}\cdot \cos \left(x\right)\right]}_0^{\pi }$

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

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

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

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

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

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

= $e^{3\pi }+0$

= ${\left[-\frac{1}{3}{\left(3x-4\right)}^3+\frac{5}{2}{x}^2\right]}_1^2$

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

= $-\frac{1}{3}\cdot {\left(6-4\right)}^3+\frac{5}{2}\cdot 4-\left(-\frac{1}{3}\cdot {\left(3-4\right)}^3+\frac{5}{2}\cdot 1\right)$

= $-\frac{1}{3}\cdot 2^3+10-\left(-\frac{1}{3}\cdot {\left(-1\right)}^3+\frac{5}{2}\right)$

= $-\frac{1}{3}\cdot 8+10-\left(-\frac{1}{3}\cdot \left(-1\right)+\frac{5}{2}\right)$

= $-\frac{8}{3}+10-\left(\frac{1}{3}+\frac{5}{2}\right)$

= $-\frac{8}{3}+\frac{30}{3}-\left(\frac{2}{6}+\frac{15}{6}\right)$

= $\frac{22}{3}-1·\frac{17}{6}$

= $\frac{22}{3}-\frac{17}{6}$

= $\frac{44}{6}-\frac{17}{6}$

= $\frac{9}{2}$