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

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

Somit gilt:

$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = -2 +6 = 4

= ${\left[-\frac{5}{3}{x}^3-\frac{1}{2}{x}^2\right]}_1^5$

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

= $-\frac{5}{3}\cdot 125-\frac{1}{2}\cdot 25-\left(-\frac{5}{3}\cdot 1-\frac{1}{2}\cdot 1\right)$

= $-\frac{625}{3}-\frac{25}{2}-\left(-\frac{5}{3}-\frac{1}{2}\right)$

= $-\frac{1250}{6}-\frac{75}{6}-\left(-\frac{10}{6}-\frac{3}{6}\right)$

= $-\frac{1325}{6}-1·\left(-\frac{13}{6}\right)$

= $-\frac{1325}{6}+\frac{13}{6}$

= $-\frac{656}{3}$

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

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

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

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

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

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

= $-3$

= ${\left[-\frac{1}{4}{\left(-2x+5\right)}^4\right]}_2^5$

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

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

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

= $-\frac{1}{4}\cdot 625+\frac{1}{4}\cdot 1$

= $-\frac{625}{4}+\frac{1}{4}$

= $-156$

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

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

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

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

= $\frac{4}{3}{e}^{\frac{9}{2}\pi }-\left(\frac{8}{6}-\frac{3}{6}\right)$

= $\frac{4}{3}{e}^{\frac{9}{2}\pi }-1·\frac{5}{6}$

= $\frac{4}{3}{e}^{\frac{9}{2}\pi }-\frac{5}{6}$

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

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

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

= $0-0$

= 0