$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 3 = 2 ⋅ 3 = 6.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 6 +3 -1.5 = 7.5

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

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

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

= $18-9-\left(\frac{2}{3}-1\right)$

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

= $9-1·\left(-\frac{1}{3}\right)$

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

= $\frac{27}{3}+\frac{1}{3}$

= $\frac{28}{3}$

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

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

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

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

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

= $-\frac{3}{2}{e}^{3\pi }+\frac{3}{2}{e}^{\frac{3}{2}\pi }+2$

= ${\left[-3e^{x-1}\right]}_1^4$

$=$ $-3e^{4-1}+3e^{1-1}$

= $-3e^3+3e^0$

= $-3e^3+3$

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

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

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

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

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

= $\frac{3}{2}+\frac{32}{81{\pi }^3}-\left(-\frac{3}{2}+\frac{32}{3{\pi }^3}\right)$

= $\frac{3}{2}+\frac{32}{81{\pi }^3}-1·\left(-\frac{3}{2}\right)-1·\frac{32}{3{\pi }^3}$

= $\frac{3}{2}+\frac{32}{81{\pi }^3}+\frac{3}{2}-\frac{32}{3{\pi }^3}$

= $\frac{3}{2}+\frac{3}{2}+\frac{32}{81{\pi }^3}-\frac{32}{3{\pi }^3}$

= $3-\frac{832}{81{\pi }^3}$

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

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

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

= $-0+1$

= $1$