$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ $\frac{\mathrm{4\; +\; <mn>3</mn>}}{2}$ = 3 ⋅ 3.5 = 10.5.

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 +10.5 = 24.5

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

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

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

= $-\frac{32}{3}-8-\left(\frac{32}{3}-8\right)$

= $-\frac{32}{3}-\frac{24}{3}-\left(\frac{32}{3}-\frac{24}{3}\right)$

= $-\frac{56}{3}-1·\frac{8}{3}$

= $-\frac{56}{3}-\frac{8}{3}$

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

= ${\left[-\frac{5}{6}{e}^{2x}-4\cdot \cos \left(x\right)\right]}_0^{\pi }$

$=$ $-\frac{5}{6}{e}^{2\cdot \pi }-4\cdot \cos \left(\pi \right)-\left(-\frac{5}{6}{e}^{2\cdot \left(0\right)}-4\cdot \cos \left(0\right)\right)$

= $-\frac{5}{6}{e}^{2\cdot \pi }-4\cdot \left(-1\right)-\left(-\frac{5}{6}{e}^0-4\cdot 1\right)$

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

= $-\frac{5}{6}{e}^{2\cdot \pi }+4-\left(-\frac{5}{6}-\frac{24}{6}\right)$

= $-\frac{5}{6}{e}^{2\cdot \pi }+4-1·\left(-\frac{29}{6}\right)$

= $-\frac{5}{6}{e}^{2\cdot \pi }+4+\frac{29}{6}$

= $-\frac{5}{6}{e}^{2\pi }+\frac{53}{6}$

= ${\left[-\frac{1}{6}{\left(2x-4\right)}^3-3x\right]}_0^1$

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

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

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

= $-\frac{1}{6}\cdot \left(-8\right)-3-\left(-\frac{1}{6}\cdot \left(-64\right)+0\right)$

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

= $\frac{4}{3}-\frac{9}{3}-\left(\frac{32}{3}+0\right)$

= $-\frac{5}{3}-\frac{32}{3}$

= $-\frac{37}{3}$

= ${\left[2\cdot \cos \left(x\right)-\frac{9}{2}\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $2\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{9}{2}\cdot \sin \left(\frac{1}{2}\pi \right)-\left(2\cdot \cos \left(0\right)-\frac{9}{2}\cdot \sin \left(0\right)\right)$

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

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

= $0-\frac{9}{2}-2$

= $-\frac{9}{2}-2$

= $-\frac{9}{2}-\frac{4}{2}$

= $-\frac{13}{2}$

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

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

= $\cos \left(-6\pi \right)-\cos \left(-3\pi \right)$

= $1-\left(-1\right)$

= $1+1$

= $2$