$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{2}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -3 +2 +4 = 3

= ${\left[-\frac{5}{2}{x}^2-3x\right]}_1^4$

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

= $-\frac{5}{2}\cdot 16-12-\left(-\frac{5}{2}\cdot 1-3\right)$

= $-40-12-\left(-\frac{5}{2}-3\right)$

= $-52-\left(-\frac{5}{2}-\frac{6}{2}\right)$

= $-52-1·\left(-\frac{11}{2}\right)$

= $-52+\frac{11}{2}$

= $-\frac{104}{2}+\frac{11}{2}$

= $-\frac{93}{2}$

= ${\left[\frac{5}{3}{x}^{\frac{3}{2}}-\frac{3}{4}{x}^4\right]}_4^{16}$

= ${\left[\frac{5}{3}{\left(\sqrt{x}\right)}^3-\frac{3}{4}{x}^4\right]}_4^{16}$

$=$ $\frac{5}{3}{\left(\sqrt{16}\right)}^3-\frac{3}{4}\cdot {16}^4-\left(\frac{5}{3}{\left(\sqrt{4}\right)}^3-\frac{3}{4}\cdot 4^4\right)$

= $\frac{5}{3}\cdot 4^3-\frac{3}{4}\cdot 65536-\left(\frac{5}{3}\cdot 2^3-\frac{3}{4}\cdot 256\right)$

= $\frac{5}{3}\cdot 64-49152-\left(\frac{5}{3}\cdot 8-192\right)$

= $\frac{320}{3}-49152-\left(\frac{40}{3}-192\right)$

= $\frac{320}{3}-\frac{147456}{3}-\left(\frac{40}{3}-\frac{576}{3}\right)$

= $-\frac{147136}{3}-1·\left(-\frac{536}{3}\right)$

= $-\frac{147136}{3}+\frac{536}{3}$

= $-\frac{146600}{3}$

= ${\left[-\frac{1}{6}{\left(2x-4\right)}^3\right]}_2^3$

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

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

= $-\frac{1}{6}\cdot 2^3+\frac{1}{6}\cdot 0^3$

= $-\frac{1}{6}\cdot 8+\frac{1}{6}\cdot 0$

= $-\frac{4}{3}+0$

= $-\frac{4}{3}+0$

= $-\frac{4}{3}$

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

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

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

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

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

= $-e^{3\pi }+1+e^{\pi }-1·\left(-1\right)$

= $-e^{3\pi }+1+e^{\pi }+1$

= $-e^{3\pi }+e^{\pi }+1+1$

= $-e^{3\pi }+e^{\pi }+2$

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

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

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

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

= $-\frac{1}{2}\cdot 125+\frac{1}{2}\cdot 1$

= $-\frac{125}{2}+\frac{1}{2}$

= $-62$