$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 2 -6 -12 = -16

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

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

= $-4+8-\left(-9+12\right)$

= $-4+8-1·\left(-9\right)-1·12$

= $-4+8+9-12$

= $1$

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

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

= $-\frac{2}{5}\cdot 3125+\frac{5}{4}{e}^{-10}-\left(-\frac{2}{5}\cdot 1+\frac{5}{4}{e}^{-2}\right)$

= $-1250+\frac{5}{4}{e}^{-10}-\left(-\frac{2}{5}+\frac{5}{4}{e}^{-2}\right)$

= $\frac{5}{4}{e}^{-10}-1250-\left(\frac{5}{4}{e}^{-2}-\frac{2}{5}\right)$

= $-\frac{5}{4}{e}^{-2}-1·\left(-\frac{2}{5}\right)+\frac{5}{4}{e}^{-10}-1250$

= $-\frac{5}{4}{e}^{-2}+\frac{2}{5}+\frac{5}{4}{e}^{-10}-1250$

= $-\frac{5}{4}{e}^{-2}+\frac{5}{4}{e}^{-10}+\frac{2}{5}-1250$

= $-\frac{5}{4}{e}^{-2}+\frac{5}{4}{e}^{-10}-\frac{6248}{5}$

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

$=$ $-e^{4-3}+e^{1-3}$

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

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

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

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

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

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

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

= $2e^{-\frac{3}{2}\pi }-2-2e^{-\frac{1}{2}\pi }-1·2$

= $2e^{-\frac{3}{2}\pi }-2-2e^{-\frac{1}{2}\pi }-2$

= $-2e^{-\frac{1}{2}\pi }+2e^{-\frac{3}{2}\pi }-2-2$

= $-2e^{-\frac{1}{2}\pi }+2e^{-\frac{3}{2}\pi }-4$

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

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

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

= $-2\cdot 0+2\cdot 1$

= $0+2$

= $2$