$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 1 = 2 ⋅ 1 = 2.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₅ = (10 - 8) ⋅ $\frac{\mathrm{1\; +\; <mn>2</mn>}}{2}$ = 2 ⋅ 1.5 = 3.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ + I₅ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 -2 +1 +2 +3 = 0

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

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

= $\frac{5}{3}\cdot 125+\frac{5}{2}\cdot 25-\left(\frac{5}{3}\cdot 1+\frac{5}{2}\cdot 1\right)$

= $\frac{625}{3}+\frac{125}{2}-\left(\frac{5}{3}+\frac{5}{2}\right)$

= $\frac{1250}{6}+\frac{375}{6}-\left(\frac{10}{6}+\frac{15}{6}\right)$

= $\frac{1625}{6}-1·\frac{25}{6}$

= $\frac{1625}{6}-\frac{25}{6}$

= $\frac{800}{3}$

= ${\left[\frac{2}{3}{x}^6-\frac{7}{2}\cdot \sin \left(x\right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $\frac{2}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{7}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(\frac{2}{3}\cdot {\left(0\right)}^6-\frac{7}{2}\cdot \sin \left(0\right)\right)$

= $\frac{2}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6-\frac{7}{2}\cdot \left(-1\right)-\left(\frac{2}{3}\cdot 0-\frac{7}{2}\cdot 0\right)$

= $\frac{2}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6+\frac{7}{2}-\left(0+0\right)$

= $\frac{7}{2}+\frac{243}{32}\cdot {\pi }^6+0$

= $\frac{7}{2}+\frac{243}{32}\cdot {\pi }^6$

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

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

= $-\cos \left(-2\pi \right)+\cos (-\pi )$

= $-1-1$

= $-2$

= ${\left[x^3+\frac{5}{3}{x}^{\frac{3}{2}}\right]}_4^{16}$

= ${\left[x^3+\frac{5}{3}{\left(\sqrt{x}\right)}^3\right]}_4^{16}$

$=$ ${16}^3+\frac{5}{3}{\left(\sqrt{16}\right)}^3-\left(4^3+\frac{5}{3}{\left(\sqrt{4}\right)}^3\right)$

= $4096+\frac{5}{3}\cdot 4^3-\left(64+\frac{5}{3}\cdot 2^3\right)$

= $4096+\frac{5}{3}\cdot 64-\left(64+\frac{5}{3}\cdot 8\right)$

= $4096+\frac{320}{3}-\left(64+\frac{40}{3}\right)$

= $\frac{12288}{3}+\frac{320}{3}-\left(\frac{192}{3}+\frac{40}{3}\right)$

= $\frac{12608}{3}-1·\frac{232}{3}$

= $\frac{12608}{3}-\frac{232}{3}$

= $\frac{12376}{3}$

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

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

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

= $-\frac{1}{3}\cdot \left(-1\right)+8-\left(-\frac{1}{3}\cdot \left(-27\right)+0\right)$

= $\frac{1}{3}+8-\left(9+0\right)$

= $\frac{1}{3}+\frac{24}{3}-9$

= $\frac{25}{3}-9$

= $\frac{25}{3}-\frac{27}{3}$

= $-\frac{2}{3}$