$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 7) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 2 -4.5 -9 = -11.5

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

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

= $\frac{3}{2}\cdot 1-4-\left(\frac{3}{2}\cdot 9+12\right)$

= $\frac{3}{2}-4-\left(\frac{27}{2}+12\right)$

= $\frac{3}{2}-\frac{8}{2}-\left(\frac{27}{2}+\frac{24}{2}\right)$

= $-\frac{5}{2}-1·\frac{51}{2}$

= $-\frac{5}{2}-\frac{51}{2}$

= $-28$

= ${\left[\frac{1}{4}{x}^6+\frac{8}{3}{x}^{\frac{3}{2}}\right]}_0^1$

= ${\left[\frac{1}{4}{x}^6+\frac{8}{3}{\left(\sqrt{x}\right)}^3\right]}_0^1$

$=$ $\frac{1}{4}\cdot 1^6+\frac{8}{3}{\left(\sqrt{1}\right)}^3-\left(\frac{1}{4}\cdot 0^6+\frac{8}{3}{\left(\sqrt{0}\right)}^3\right)$

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

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

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

= $\frac{3}{12}+\frac{32}{12}+0$

= $\frac{35}{12}$

= ${\left[\frac{1}{3}{\left(-3x+7\right)}^3+3x^2\right]}_2^4$

$=$ $\frac{1}{3}\cdot {\left(-3\cdot 4+7\right)}^3+3\cdot 4^2-\left(\frac{1}{3}\cdot {\left(-3\cdot 2+7\right)}^3+3\cdot 2^2\right)$

= $\frac{1}{3}\cdot {\left(-12+7\right)}^3+3\cdot 16-\left(\frac{1}{3}\cdot {\left(-6+7\right)}^3+3\cdot 4\right)$

= $\frac{1}{3}\cdot {\left(-5\right)}^3+48-\left(\frac{1}{3}\cdot 1^3+12\right)$

= $\frac{1}{3}\cdot \left(-125\right)+48-\left(\frac{1}{3}\cdot 1+12\right)$

= $-\frac{125}{3}+48-\left(\frac{1}{3}+12\right)$

= $-\frac{125}{3}+\frac{144}{3}-\left(\frac{1}{3}+\frac{36}{3}\right)$

= $\frac{19}{3}-1·\frac{37}{3}$

= $\frac{19}{3}-\frac{37}{3}$

= $-6$

= ${\left[-\frac{5}{6}{x}^{-2}+\frac{3}{4}\cdot \sin \left(x\right)\right]}_{\pi }^{\frac{3}{2}\pi }$

= ${\left[-\frac{5}{6x^2}+\frac{3}{4}\cdot \sin \left(x\right)\right]}_{\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{5}{6{\left(\frac{3}{2}\pi \right)}^2}+\frac{3}{4}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{5}{6{\pi }^2}+\frac{3}{4}\cdot \sin \left(\pi \right)\right)$

= $-\frac{5}{6{\left(\frac{3}{2}\pi \right)}^2}+\frac{3}{4}\cdot \left(-1\right)-\left(-\frac{5}{6{\pi }^2}+\frac{3}{4}\cdot 0\right)$

= $-\frac{5}{6{\left(\frac{3}{2}\pi \right)}^2}-\frac{3}{4}-\left(-\frac{5}{6{\pi }^2}+0\right)$

= $-\frac{3}{4}-\frac{10}{27{\pi }^2}+\frac{5}{6{\pi }^2}$

= $-\frac{3}{4}+\frac{25}{54{\pi }^2}$

= ${\left[-\frac{1}{8}{\left(2x-2\right)}^4\right]}_0^2$

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

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

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

= $-\frac{1}{8}\cdot 16+\frac{1}{8}\cdot 16$

= $-2+2$

= 0