$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

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.

Somit gilt:

$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = -4 +4.5 = 0.5

= ${\left[-\frac{4}{3}{x}^3+4x\right]}_1^4$

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

= $-\frac{4}{3}\cdot 64+16-\left(-\frac{4}{3}\cdot 1+4\right)$

= $-\frac{256}{3}+16-\left(-\frac{4}{3}+4\right)$

= $-\frac{256}{3}+\frac{48}{3}-\left(-\frac{4}{3}+\frac{12}{3}\right)$

= $-\frac{208}{3}-1·\frac{8}{3}$

= $-\frac{208}{3}-\frac{8}{3}$

= $-72$

= ${\left[\frac{9}{4}\cdot \cos \left(x\right)-\frac{1}{3}{x}^{-3}\right]}_{\pi }^{2\pi }$

= ${\left[\frac{9}{4}\cdot \cos \left(x\right)-\frac{1}{3x^3}\right]}_{\pi }^{2\pi }$

$=$ $\frac{9}{4}\cdot \cos \left(2\pi \right)-\frac{1}{3{\left(2\pi \right)}^3}-\left(\frac{9}{4}\cdot \cos \left(\pi \right)-\frac{1}{3{\pi }^3}\right)$

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

= $\frac{9}{4}-\frac{1}{3{\left(2\pi \right)}^3}-\left(-\frac{9}{4}-\frac{1}{3{\pi }^3}\right)$

= $\frac{9}{4}-\frac{1}{24{\pi }^3}-\left(-\frac{9}{4}-\frac{1}{3{\pi }^3}\right)$

= $\frac{9}{4}-\frac{1}{24{\pi }^3}-1·\left(-\frac{9}{4}\right)-1·\left(-\frac{1}{3{\pi }^3}\right)$

= $\frac{9}{4}-\frac{1}{24{\pi }^3}+\frac{9}{4}+\frac{1}{3{\pi }^3}$

= $\frac{9}{4}+\frac{9}{4}-\frac{1}{24{\pi }^3}+\frac{1}{3{\pi }^3}$

= $\frac{9}{2}+\frac{7}{24{\pi }^3}$

= ${\left[-{\left(-2x+5\right)}^{-1}\right]}_3^4$

= ${\left[-\frac{1}{-2x+5}\right]}_3^4$

$=$ $-\frac{1}{-2\cdot 4+5}+\frac{1}{-2\cdot 3+5}$

= $-\frac{1}{-8+5}+\frac{1}{-6+5}$

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

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

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

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

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

= ${\left[2\cdot \cos \left(x\right)+\frac{1}{3}{x}^6\right]}_0^{\pi }$

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

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

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

= $-2+\frac{1}{3}\cdot {\pi }^6-2$

= $-2-2+\frac{1}{3}\cdot {\pi }^6$

= $-4+\frac{1}{3}\cdot {\pi }^6$

= ${\left[\frac{2}{3}{\left(3x-6\right)}^{-1}\right]}_3^6$

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

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

= $\frac{2}{3\left(18-6\right)}-\frac{2}{3\left(9-6\right)}$

= $\frac{2}{3\cdot 12}-\frac{2}{3\cdot 3}$

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

= $\frac{1}{18}-\frac{2}{9}$

= $\frac{1}{18}-\frac{4}{18}$

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