$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

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

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -8 -4 +1.5 = -10.5

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

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

= $1+3-\left(9+9\right)$

= $1+3-1·9-1·9$

= $1+3-9-9$

= $-14$

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

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

= $\frac{1}{3}{e}^3+2-\left(\frac{1}{3}{e}^{-9}-6\right)$

= $\frac{1}{3}{e}^3+2-\frac{1}{3}{e}^{-9}-1·\left(-6\right)$

= $\frac{1}{3}{e}^3+2-\frac{1}{3}{e}^{-9}+6$

= $\frac{1}{3}{e}^3-\frac{1}{3}{e}^{-9}+2+6$

= $\frac{1}{3}{e}^3-\frac{1}{3}{e}^{-9}+8$

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

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

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

= $-\frac{2}{3}\cdot 0+\frac{2}{3}\cdot 0$

= $0+0$

= 0

= ${\left[\frac{2}{5}{x}^5+\frac{5}{4}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

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

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

= $-\frac{5}{4}+\frac{243}{80}\cdot {\pi }^5-\left(\frac{5}{4}+\frac{1}{80}\cdot {\pi }^5\right)$

= $-\frac{5}{4}+\frac{243}{80}\cdot {\pi }^5-1·\frac{5}{4}-1·\frac{1}{80}\cdot {\pi }^5$

= $-\frac{5}{4}+\frac{243}{80}\cdot {\pi }^5-\frac{5}{4}-\frac{1}{80}\cdot {\pi }^5$

= $-\frac{5}{4}-\frac{5}{4}+\frac{243}{80}\cdot {\pi }^5-\frac{1}{80}\cdot {\pi }^5$

= $-\frac{5}{2}+\frac{121}{40}\cdot {\pi }^5$

= ${\left[\frac{2}{3}\ln \left(\left|$ -3x+6$\right|\right)\right]}_3^5$

$=$ $\frac{2}{3}\ln \left(\left|$ -3\cdot 5+6$\right|\right)-\frac{2}{3}\ln \left(\left|$ -3\cdot 3+6$\right|\right)$

= $\frac{2}{3}\ln \left(\left|$ -15+6$\right|\right)-\frac{2}{3}\ln \left(\left|$ -9+6$\right|\right)$

= $\frac{2}{3}\ln \left(9\right)-\frac{2}{3}\ln \left(\left|$ -9+6$\right|\right)$

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