$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -4 -2 +6 = 0

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

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

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

= $\frac{27}{2}+3-\left(0+0\right)$

= $\frac{27}{2}+\frac{6}{2}+0$

= $\frac{33}{2}$

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

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

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

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

= $\frac{1}{4}\cdot {\pi }^4+0$

= $\frac{1}{4}\cdot {\pi }^4$

= ${\left[-\frac{3}{4}{\left(-x+2\right)}^4\right]}_2^5$

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

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

= $-\frac{3}{4}\cdot 81+\frac{3}{4}\cdot 0$

= $-\frac{243}{4}+0$

= $-\frac{243}{4}+0$

= $-\frac{243}{4}$

= ${\left[\frac{7}{2}\cdot \cos \left(x\right)+\frac{5}{9}{x}^{-3}\right]}_{\pi }^{\frac{3}{2}\pi }$

= ${\left[\frac{7}{2}\cdot \cos \left(x\right)+\frac{5}{9x^3}\right]}_{\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{7}{2}\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{5}{9{\left(\frac{3}{2}\pi \right)}^3}-\left(\frac{7}{2}\cdot \cos \left(\pi \right)+\frac{5}{9{\pi }^3}\right)$

= $\frac{7}{2}\cdot 0+\frac{5}{9{\left(\frac{3}{2}\pi \right)}^3}-\left(\frac{7}{2}\cdot \left(-1\right)+\frac{5}{9{\pi }^3}\right)$

= $0+\frac{5}{9{\left(\frac{3}{2}\pi \right)}^3}-\left(-\frac{7}{2}+\frac{5}{9{\pi }^3}\right)$

= $\frac{40}{243{\pi }^3}-\left(-\frac{7}{2}+\frac{5}{9{\pi }^3}\right)$

= $-1·\left(-\frac{7}{2}\right)-1·\frac{5}{9{\pi }^3}+\frac{40}{243{\pi }^3}$

= $\frac{7}{2}-\frac{5}{9{\pi }^3}+\frac{40}{243{\pi }^3}$

= $\frac{7}{2}-\frac{95}{243{\pi }^3}$

= ${\left[\frac{1}{3}{\left(-3x+6\right)}^3+3x\right]}_2^3$

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

= $\frac{1}{3}\cdot {\left(-9+6\right)}^3+9-\left(\frac{1}{3}\cdot {\left(-6+6\right)}^3+6\right)$

= $\frac{1}{3}\cdot {\left(-3\right)}^3+9-\left(\frac{1}{3}\cdot 0^3+6\right)$

= $\frac{1}{3}\cdot \left(-27\right)+9-\left(\frac{1}{3}\cdot 0+6\right)$

= $-9+9-\left(0+6\right)$

= $0-6$

= $-6$