$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 3 = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -2 +4.5 +6 = 8.5

= ${\left[\frac{2}{3}{x}^3-\frac{1}{2}{x}^2\right]}_1^2$

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

= $\frac{2}{3}\cdot 8-\frac{1}{2}\cdot 4-\left(\frac{2}{3}\cdot 1-\frac{1}{2}\cdot 1\right)$

= $\frac{16}{3}-2-\left(\frac{2}{3}-\frac{1}{2}\right)$

= $\frac{16}{3}-\frac{6}{3}-\left(\frac{4}{6}-\frac{3}{6}\right)$

= $\frac{10}{3}-1·\frac{1}{6}$

= $\frac{10}{3}-\frac{1}{6}$

= $\frac{20}{6}-\frac{1}{6}$

= $\frac{19}{6}$

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

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

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

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

= $-\frac{243}{32}\cdot {\pi }^6+1$

= ${\left[-\frac{1}{3}{\left(-x+1\right)}^3-\frac{3}{2}{x}^2\right]}_1^2$

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

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

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

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

= $\frac{1}{3}-\frac{18}{3}-\left(0-\frac{3}{2}\right)$

= $-\frac{17}{3}+\frac{3}{2}$

= $-\frac{34}{6}+\frac{9}{6}$

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

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

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

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

= $\frac{7}{3}+0-\left(0-3\right)$

= $\frac{7}{3}+0+3$

= $\frac{7}{3}+3$

= $\frac{7}{3}+\frac{9}{3}$

= $\frac{16}{3}$

= ${\left[-\frac{1}{3}{\left(2x-5\right)}^3+2x\right]}_2^5$

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

= $-\frac{1}{3}\cdot {\left(10-5\right)}^3+10-\left(-\frac{1}{3}\cdot {\left(4-5\right)}^3+4\right)$

= $-\frac{1}{3}\cdot 5^3+10-\left(-\frac{1}{3}\cdot {\left(-1\right)}^3+4\right)$

= $-\frac{1}{3}\cdot 125+10-\left(-\frac{1}{3}\cdot \left(-1\right)+4\right)$

= $-\frac{125}{3}+10-\left(\frac{1}{3}+4\right)$

= $-\frac{125}{3}+\frac{30}{3}-\left(\frac{1}{3}+\frac{12}{3}\right)$

= $-\frac{95}{3}-1·\frac{13}{3}$

= $-\frac{95}{3}-\frac{13}{3}$

= $-36$