$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

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

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -6 -4.5 +1 = -9.5

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

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

= $-\frac{1}{3}\cdot 27-\frac{3}{2}\cdot 9-\left(-\frac{1}{3}\cdot \left(-1\right)-\frac{3}{2}\cdot 1\right)$

= $-9-\frac{27}{2}-\left(\frac{1}{3}-\frac{3}{2}\right)$

= $-\frac{18}{2}-\frac{27}{2}-\left(\frac{2}{6}-\frac{9}{6}\right)$

= $-\frac{45}{2}-1·\left(-\frac{7}{6}\right)$

= $-\frac{45}{2}+\frac{7}{6}$

= $-\frac{135}{6}+\frac{7}{6}$

= $-\frac{64}{3}$

= ${\left[-\frac{1}{3}{x}^6-\frac{5}{3}{e}^{-3x}\right]}_1^4$

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

= $-\frac{1}{3}\cdot 4096-\frac{5}{3}{e}^{-12}-\left(-\frac{1}{3}\cdot 1-\frac{5}{3}{e}^{-3}\right)$

= $-\frac{4096}{3}-\frac{5}{3}{e}^{-12}-\left(-\frac{1}{3}-\frac{5}{3}{e}^{-3}\right)$

= $-\frac{5}{3}{e}^{-12}-\frac{4096}{3}-\left(-\frac{5}{3}{e}^{-3}-\frac{1}{3}\right)$

= $\frac{5}{3}{e}^{-3}-1·\left(-\frac{1}{3}\right)-\frac{5}{3}{e}^{-12}-\frac{4096}{3}$

= $\frac{5}{3}{e}^{-3}+\frac{1}{3}-\frac{5}{3}{e}^{-12}-\frac{4096}{3}$

= $\frac{5}{3}{e}^{-3}-\frac{5}{3}{e}^{-12}+\frac{1}{3}-\frac{4096}{3}$

= $\frac{5}{3}{e}^{-3}-\frac{5}{3}{e}^{-12}-1365$

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

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

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

= $2\cdot \left(-1\right)-2\cdot 0$

= $-2+0$

= $-2$

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

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

= $\frac{4}{3}\cdot 0-\frac{5}{2}\cdot {\left(\frac{3}{2}\pi \right)}^2-\left(\frac{4}{3}\cdot 0-\frac{5}{2}\cdot {\left(\frac{1}{2}\pi \right)}^2\right)$

= $0-\frac{5}{2}\cdot {\left(\frac{3}{2}\pi \right)}^2-\left(0-\frac{5}{2}\cdot {\left(\frac{1}{2}\pi \right)}^2\right)$

= $-\frac{45}{8}\cdot {\pi }^2+\frac{5}{8}\cdot {\pi }^2$

= $-5\cdot {\pi }^2$

= ${\left[-\frac{2}{3}{e}^{-3x+7}\right]}_2^4$

$=$ $-\frac{2}{3}{e}^{-3\cdot 4+7}+\frac{2}{3}{e}^{-3\cdot 2+7}$

= $-\frac{2}{3}{e}^{-12+7}+\frac{2}{3}{e}^{-6+7}$

= $-\frac{2}{3}{e}^{-5}+\frac{2}{3}e$

= $-\frac{2}{3}{e}^{-5}+\frac{2}{3}e$