$$\underset{3}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

Somit gilt:

$$\underset{3}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -3 +2 = -1

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

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

= $-2\cdot 1-5-\left(-2\cdot 1+5\right)$

= $-2-5-\left(-2+5\right)$

= $-7-1·3$

= $-7-3$

= $-10$

= ${\left[-\frac{5}{8}{x}^4+\frac{9}{2}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{5}{8}\cdot {\left(\frac{3}{2}\pi \right)}^4+\frac{9}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{5}{8}\cdot {\left(\frac{1}{2}\pi \right)}^4+\frac{9}{2}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{8}\cdot {\left(\frac{3}{2}\pi \right)}^4+\frac{9}{2}\cdot \left(-1\right)-\left(-\frac{5}{8}\cdot {\left(\frac{1}{2}\pi \right)}^4+\frac{9}{2}\cdot 1\right)$

= $-\frac{5}{8}\cdot {\left(\frac{3}{2}\pi \right)}^4-\frac{9}{2}-\left(-\frac{5}{8}\cdot {\left(\frac{1}{2}\pi \right)}^4+\frac{9}{2}\right)$

= $-\frac{9}{2}-\frac{405}{128}\cdot {\pi }^4-\left(\frac{9}{2}-\frac{5}{128}\cdot {\pi }^4\right)$

= $-\frac{9}{2}-\frac{405}{128}\cdot {\pi }^4-1·\frac{9}{2}-1·\left(-\frac{5}{128}\cdot {\pi }^4\right)$

= $-\frac{9}{2}-\frac{405}{128}\cdot {\pi }^4-\frac{9}{2}+\frac{5}{128}\cdot {\pi }^4$

= $-\frac{9}{2}-\frac{9}{2}-\frac{405}{128}\cdot {\pi }^4+\frac{5}{128}\cdot {\pi }^4$

= $-9-\frac{25}{8}\cdot {\pi }^4$

= ${\left[-\frac{1}{9}{\left(3x-7\right)}^3+\frac{3}{2}{x}^2\right]}_0^1$

$=$ $-\frac{1}{9}\cdot {\left(3\cdot 1-7\right)}^3+\frac{3}{2}\cdot 1^2-\left(-\frac{1}{9}\cdot {\left(3\cdot 0-7\right)}^3+\frac{3}{2}\cdot 0^2\right)$

= $-\frac{1}{9}\cdot {\left(3-7\right)}^3+\frac{3}{2}\cdot 1-\left(-\frac{1}{9}\cdot {\left(0-7\right)}^3+\frac{3}{2}\cdot 0\right)$

= $-\frac{1}{9}\cdot {\left(-4\right)}^3+\frac{3}{2}-\left(-\frac{1}{9}\cdot {\left(-7\right)}^3+0\right)$

= $-\frac{1}{9}\cdot \left(-64\right)+\frac{3}{2}-\left(-\frac{1}{9}\cdot \left(-343\right)+0\right)$

= $\frac{64}{9}+\frac{3}{2}-\left(\frac{343}{9}+0\right)$

= $\frac{128}{18}+\frac{27}{18}-\left(\frac{343}{9}+0\right)$

= $\frac{155}{18}-\frac{343}{9}$

= $\frac{155}{18}-\frac{686}{18}$

= $\frac{155}{18}-\frac{343}{9}$

= $\frac{155}{18}-\frac{686}{18}$

= $-\frac{59}{2}$

= ${\left[\frac{7}{3}\cdot \sin \left(x\right)-5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{7}{3}\cdot \sin \left(\pi \right)-5\cdot \cos \left(\pi \right)-\left(\frac{7}{3}\cdot \sin \left(\frac{1}{2}\pi \right)-5\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{7}{3}\cdot 0-5\cdot \left(-1\right)-\left(\frac{7}{3}\cdot 1-5\cdot 0\right)$

= $0+5-\left(\frac{7}{3}+0\right)$

= $5-\left(\frac{7}{3}+0\right)$

= $5-\frac{7}{3}$

= $\frac{15}{3}-\frac{7}{3}$

= $\frac{8}{3}$

= ${\left[e^{-3x+4}\right]}_2^4$

$=$ $e^{-3\cdot 4+4}-e^{-3\cdot 2+4}$

= $e^{-12+4}-e^{-6+4}$

= $e^{-8}-e^{-2}$