$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

Somit gilt:

$$\underset{0}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = -2 +1.5 = -0.5

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

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

= $-\frac{5}{2}\cdot 1-3-\left(-\frac{5}{2}\cdot 4-6\right)$

= $-\frac{5}{2}-3-\left(-10-6\right)$

= $-\frac{5}{2}-\frac{6}{2}-1·\left(-16\right)$

= $-\frac{11}{2}+16$

= $-\frac{11}{2}+\frac{32}{2}$

= $\frac{21}{2}$

= ${\left[\frac{1}{3}{x}^6+8\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{1}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6+8\cdot \sin \left(\frac{3}{2}\pi \right)-\left(\frac{1}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6+8\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $\frac{1}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6+8\cdot \left(-1\right)-\left(\frac{1}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6+8\cdot 1\right)$

= $\frac{1}{3}\cdot {\left(\frac{3}{2}\pi \right)}^6-8-\left(\frac{1}{3}\cdot {\left(\frac{1}{2}\pi \right)}^6+8\right)$

= $-8+\frac{243}{64}\cdot {\pi }^6-\left(8+\frac{1}{192}\cdot {\pi }^6\right)$

= $-8+\frac{243}{64}\cdot {\pi }^6-1·8-1·\frac{1}{192}\cdot {\pi }^6$

= $-8+\frac{243}{64}\cdot {\pi }^6-8-\frac{1}{192}\cdot {\pi }^6$

= $-8-8+\frac{243}{64}\cdot {\pi }^6-\frac{1}{192}\cdot {\pi }^6$

= $-16+\frac{91}{24}\cdot {\pi }^6$

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

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

= $-e^{-1}+e^2$

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

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

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

= $-\frac{243}{64}\cdot {\pi }^4+\frac{3}{64}\cdot {\pi }^4$

= $-\frac{15}{4}\cdot {\pi }^4$

= ${\left[\frac{1}{12}{\left(3x-3\right)}^4+\frac{5}{2}{x}^2\right]}_2^5$

$=$ $\frac{1}{12}\cdot {\left(3\cdot 5-3\right)}^4+\frac{5}{2}\cdot 5^2-\left(\frac{1}{12}\cdot {\left(3\cdot 2-3\right)}^4+\frac{5}{2}\cdot 2^2\right)$

= $\frac{1}{12}\cdot {\left(15-3\right)}^4+\frac{5}{2}\cdot 25-\left(\frac{1}{12}\cdot {\left(6-3\right)}^4+\frac{5}{2}\cdot 4\right)$

= $\frac{1}{12}\cdot {12}^4+\frac{125}{2}-\left(\frac{1}{12}\cdot 3^4+10\right)$

= $\frac{1}{12}\cdot 20736+\frac{125}{2}-\left(\frac{1}{12}\cdot 81+10\right)$

= $1728+\frac{125}{2}-\left(\frac{27}{4}+10\right)$

= $\frac{3456}{2}+\frac{125}{2}-\left(\frac{27}{4}+\frac{40}{4}\right)$

= $\frac{3581}{2}-1·\frac{67}{4}$

= $\frac{3581}{2}-\frac{67}{4}$

= $\frac{7162}{4}-\frac{67}{4}$

= $\frac{7095}{4}$