$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 1 = 2 ⋅ 1 = 2.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{1\; +\; <mn>4</mn>}}{2}$ = 3 ⋅ 2.5 = 7.5.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 1.5 +2 +7.5 = 11

= ${\left[\frac{5}{3}{x}^3-\frac{1}{2}{x}^2\right]}_{-2}^0$

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

= $\frac{5}{3}\cdot 0-\frac{1}{2}\cdot 0-\left(\frac{5}{3}\cdot \left(-8\right)-\frac{1}{2}\cdot 4\right)$

= $0+0-\left(-\frac{40}{3}-2\right)$

= $0-\left(-\frac{40}{3}-\frac{6}{3}\right)$

= $-1·\left(-\frac{46}{3}\right)$

= $\frac{46}{3}$

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

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

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

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

= $-\frac{4}{3}\cdot 8+\frac{2}{3}\cdot \sin \left(4\right)-\left(-\frac{4}{3}\cdot 1+\frac{2}{3}\cdot \sin \left(1\right)\right)$

= $-\frac{32}{3}+\frac{2}{3}\cdot \sin \left(4\right)-\left(-\frac{4}{3}+\frac{2}{3}\cdot \sin \left(1\right)\right)$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{32}{3}-\left(\frac{2}{3}\cdot \sin \left(1\right)-\frac{4}{3}\right)$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{32}{3}-1·\frac{2}{3}\cdot \sin \left(1\right)-1·\left(-\frac{4}{3}\right)$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{32}{3}-\frac{2}{3}\cdot \sin \left(1\right)+\frac{4}{3}$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{2}{3}\cdot \sin \left(1\right)-\frac{32}{3}+\frac{4}{3}$

= $\frac{2}{3}\cdot \sin \left(4\right)-\frac{2}{3}\cdot \sin \left(1\right)-\frac{28}{3}$

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

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

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

= $2\cdot \left(-1\right)-2\cdot 0$

= $-2+0$

= $-2$

= ${\left[4\cdot \cos \left(x\right)+\frac{3}{2}{x}^2\right]}_0^{\frac{3}{2}\pi }$

$=$ $4\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{3}{2}\cdot {\left(\frac{3}{2}\pi \right)}^2-\left(4\cdot \cos \left(0\right)+\frac{3}{2}\cdot {\left(0\right)}^2\right)$

= $4\cdot 0+\frac{3}{2}\cdot {\left(\frac{3}{2}\pi \right)}^2-\left(4\cdot 1+\frac{3}{2}\cdot 0\right)$

= $0+\frac{3}{2}\cdot {\left(\frac{3}{2}\pi \right)}^2-\left(4+0\right)$

= $\frac{27}{8}\cdot {\pi }^2-4$

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

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

= $\frac{1}{4}\cdot {\left(-8+3\right)}^4+12-\left(\frac{1}{4}\cdot {\left(-2+3\right)}^4+3\right)$

= $\frac{1}{4}\cdot {\left(-5\right)}^4+12-\left(\frac{1}{4}\cdot 1^4+3\right)$

= $\frac{1}{4}\cdot 625+12-\left(\frac{1}{4}\cdot 1+3\right)$

= $\frac{625}{4}+12-\left(\frac{1}{4}+3\right)$

= $\frac{625}{4}+\frac{48}{4}-\left(\frac{1}{4}+\frac{12}{4}\right)$

= $\frac{673}{4}-1·\frac{13}{4}$

= $\frac{673}{4}-\frac{13}{4}$

= $165$