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

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 3 = 2 ⋅ 3 = 6.

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 +3 +6 = 6

= ${\left[-x^3-4x\right]}_{-2}^{-1}$

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

= $-\left(-1\right)+4-\left(-\left(-8\right)+8\right)$

= $1+4-\left(8+8\right)$

= $5-1·16$

= $5-16$

= $-11$

= ${\left[\frac{1}{2}{x}^{\frac{3}{2}}-e^{2x}\right]}_1^{16}$

= ${\left[\frac{1}{2}{\left(\sqrt{x}\right)}^3-e^{2x}\right]}_1^{16}$

$=$ $\frac{1}{2}{\left(\sqrt{16}\right)}^3-e^{2\cdot 16}-\left(\frac{1}{2}{\left(\sqrt{1}\right)}^3-e^{2\cdot 1}\right)$

= $\frac{1}{2}\cdot 4^3-e^{32}-\left(\frac{1}{2}\cdot 1^3-e^2\right)$

= $\frac{1}{2}\cdot 64-e^{32}-\left(\frac{1}{2}\cdot 1-e^2\right)$

= $32-e^{32}-\left(\frac{1}{2}-e^2\right)$

= $-e^{32}+32-\left(-e^2+\frac{1}{2}\right)$

= $-e^{32}+32+e^2-1·\frac{1}{2}$

= $-e^{32}+32+e^2-\frac{1}{2}$

= $-e^{32}+e^2+32-\frac{1}{2}$

= $-e^{32}+e^2+\frac{63}{2}$

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

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

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

= $\frac{1}{2}\cdot 0-\frac{1}{2}\cdot 0$

= $0+0$

= 0

= ${\left[\frac{4}{15}{x}^{\frac{5}{4}}+\frac{4}{3}{x}^3\right]}_4^9$

= ${\left[\frac{4}{15}{\left(\sqrt[4]{x}\right)}^5+\frac{4}{3}{x}^3\right]}_4^9$

$=$ $\frac{4}{15}{\left(\sqrt[4]{9}\right)}^5+\frac{4}{3}\cdot 9^3-\left(\frac{4}{15}{\left(\sqrt[4]{4}\right)}^5+\frac{4}{3}\cdot 4^3\right)$

= $\frac{4}{15}\cdot 1^5+\frac{4}{3}\cdot 729-\left(\frac{4}{15}\cdot 1^5+\frac{4}{3}\cdot 64\right)$

= $\frac{4}{15}\cdot 1+972-\left(\frac{4}{15}\cdot 1+\frac{256}{3}\right)$

= $\frac{4}{15}+972-\left(\frac{4}{15}+\frac{256}{3}\right)$

= $\frac{4}{15}+\frac{14580}{15}-\left(\frac{4}{15}+\frac{1280}{15}\right)$

= $\frac{14584}{15}-1·\frac{428}{5}$

= $\frac{14584}{15}-\frac{428}{5}$

= $\frac{14584}{15}-\frac{1284}{15}$

= $\frac{14584}{15}-\frac{428}{5}$

= $\frac{14584}{15}-\frac{1284}{15}$

= $\frac{2660}{3}$

= ${\left[-e^{-x+1}\right]}_0^3$

$=$ $-e^{-3+1}+e^{-0+1}$

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

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