$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ gibt des orientierten (also mit Vorzeichen behafteten) Flächeninhalt zwischen dem Graph der Funktion und der x-Achse an.

Um diesen aus dem Schaubild zu berechnen unterteilen wir die Fläche in Rechtecke, Parallelogrammen und ggf. Dreiecke:

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{-3\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 3.5 ) = -7.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 -9 -7 = -19

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

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

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

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

= $\mathrm{2,5}+5-\left(\mathrm{2,5}-5\right)$

= $\mathrm{7,5}-1·\left(-\mathrm{2,5}\right)$

= $\mathrm{7,5}+\mathrm{2,5}$

= $10$

= ${\left[-\frac{1}{5}{x}^5-8x^{-1}\right]}_1^3$

= ${\left[-\frac{1}{5}{x}^5-\frac{8}{x}\right]}_1^3$

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

= $-\frac{1}{5}\cdot 243-8\cdot \left(\frac{1}{3}\right)-\left(-\frac{1}{5}\cdot 1-8\cdot 1\right)$

= $-\frac{243}{5}-\frac{8}{3}-\left(-\frac{1}{5}-8\right)$

= $-\frac{729}{15}-\frac{40}{15}-\left(-\frac{1}{5}-\frac{40}{5}\right)$

= $-\frac{769}{15}-1·\left(-\frac{41}{5}\right)$

= $-\frac{769}{15}+\frac{41}{5}$

= $-\frac{769}{15}+\frac{123}{15}$

= $-\frac{646}{15}$

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

$=$ $-3e^{-3+1}+3e^{-1+1}$

= $-3e^{-2}+3e^0$

= $-3e^{-2}+3$

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

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

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

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

= $3-\frac{9}{4}\cdot {\pi }^3+0$

= $3-\frac{9}{4}\cdot {\pi }^3$

= ${\left[\frac{1}{6}{\left(2x-4\right)}^3-x^2\right]}_2^4$

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

= $\frac{1}{6}\cdot {\left(8-4\right)}^3-16-\left(\frac{1}{6}\cdot {\left(4-4\right)}^3-4\right)$

= $\frac{1}{6}\cdot 4^3-16-\left(\frac{1}{6}\cdot 0^3-4\right)$

= $\frac{1}{6}\cdot 64-16-\left(\frac{1}{6}\cdot 0-4\right)$

= $\frac{32}{3}-16-\left(0-4\right)$

= $\frac{32}{3}-\frac{48}{3}+4$

= $-\frac{16}{3}+4$

= $-\frac{16}{3}+\frac{12}{3}$

= $-\frac{16}{3}+4$

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