$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 4 = 2 ⋅ 4 = 8.

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

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 6 +8 +7.5 = 21.5

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

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

= $\frac{5}{3}\cdot 0-\frac{1}{2}\cdot 0-\left(\frac{5}{3}\cdot \left(-27\right)-\frac{1}{2}\cdot 9\right)$

= $0+0-\left(-45-\frac{9}{2}\right)$

= $0-\left(-\frac{90}{2}-\frac{9}{2}\right)$

= $-1·\left(-\frac{99}{2}\right)$

= $\frac{99}{2}$

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

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

= $\frac{3}{4}\cdot {\pi }^6-\frac{1}{3}\cdot \left(-1\right)-\left(\frac{3}{4}\cdot {\left(\frac{1}{2}\pi \right)}^6-\frac{1}{3}\cdot 0\right)$

= $\frac{3}{4}\cdot {\pi }^6+\frac{1}{3}-\left(\frac{3}{4}\cdot {\left(\frac{1}{2}\pi \right)}^6+0\right)$

= $\frac{1}{3}+\frac{3}{4}\cdot {\pi }^6-\frac{3}{256}\cdot {\pi }^6$

= $\frac{1}{3}+\frac{189}{256}\cdot {\pi }^6$

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

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

= $\frac{1}{3}{e}^{6-3}-\frac{1}{3}{e}^{3-3}$

= $\frac{1}{3}{e}^3-\frac{1}{3}{e}^0$

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

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

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

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

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

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

= $\frac{1}{24{\pi }^3}-\left(-2+\frac{8}{3{\pi }^3}\right)$

= $-1·\left(-2\right)-1·\frac{8}{3{\pi }^3}+\frac{1}{24{\pi }^3}$

= $2-\frac{8}{3{\pi }^3}+\frac{1}{24{\pi }^3}$

= $2-\frac{21}{8{\pi }^3}$

= ${\left[-\frac{1}{3}{\left(3x-4\right)}^3\right]}_0^2$

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

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

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

= $-\frac{1}{3}\cdot 8+\frac{1}{3}\cdot \left(-64\right)$

= $-\frac{8}{3}-\frac{64}{3}$

= $-24$