$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

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

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

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

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

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 6 +3 -2 -4 = 3

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

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

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

= $\frac{2}{3}+\frac{3}{2}-\left(-\frac{16}{3}+6\right)$

= $\frac{4}{6}+\frac{9}{6}-\left(-\frac{16}{3}+\frac{18}{3}\right)$

= $\frac{13}{6}-1·\frac{2}{3}$

= $\frac{13}{6}-\frac{2}{3}$

= $\frac{13}{6}-\frac{4}{6}$

= $\frac{3}{2}$

= ${\left[2e^{-2x}-\frac{1}{2}{x}^6\right]}_{-1}^1$

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

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

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

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

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

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

= $-2e^2+2e^{-2}+0$

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

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

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

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

= $0+0$

= 0

= ${\left[2e^x+8\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $2e^{\frac{1}{2}\pi }+8\cdot \cos \left(\frac{1}{2}\pi \right)-\left(2e^0+8\cdot \cos \left(0\right)\right)$

= $2e^{\frac{1}{2}\pi }+8\cdot 0-\left(2+8\cdot 1\right)$

= $2e^{\frac{1}{2}\pi }+0-\left(2+8\right)$

= $2e^{\frac{1}{2}\pi }-1·10$

= $2e^{\frac{1}{2}\pi }-10$

= ${\left[{\left(-2x+5\right)}^{\frac{3}{2}}\right]}_{-10}^{-\frac{11}{2}}$

= ${\left[{\left(\sqrt{-2x+5}\right)}^3\right]}_{-10}^{-\frac{11}{2}}$

$=$ ${\left(\sqrt{-2\cdot \left(-\frac{11}{2}\right)+5}\right)}^3-{\left(\sqrt{-2\cdot \left(-10\right)+5}\right)}^3$

= ${\left(\sqrt{11+5}\right)}^3-{\left(\sqrt{20+5}\right)}^3$

= ${\left(\sqrt{16}\right)}^3-{\left(\sqrt{25}\right)}^3$

= $4^3-5^3$

= $64-125$

= $-61$