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

= ${\left[-\frac{5}{2}{x}^2-2x\right]}_0^2$

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

= $-\frac{5}{2}\cdot 4-4-\left(-\frac{5}{2}\cdot 0+0\right)$

= $-10-4-\left(0+0\right)$

= $-14+0$

= $-14$

= ${\left[7x^{-1}+\frac{5}{4}{e}^{-x}\right]}_1^5$

= ${\left[\frac{7}{x}+\frac{5}{4}{e}^{-x}\right]}_1^5$

$=$ $\frac{7}{5}+\frac{5}{4}{e}^{-5}-\left(\frac{7}{1}+\frac{5}{4}{e}^{-1}\right)$

= $7\cdot \left(\frac{1}{5}\right)+\frac{5}{4}{e}^{-5}-\left(7\cdot 1+\frac{5}{4}{e}^{-1}\right)$

= $\frac{7}{5}+\frac{5}{4}{e}^{-5}-\left(7+\frac{5}{4}{e}^{-1}\right)$

= $\frac{5}{4}{e}^{-5}+\frac{7}{5}-\left(\frac{5}{4}{e}^{-1}+7\right)$

= $-\frac{5}{4}{e}^{-1}-1·7+\frac{5}{4}{e}^{-5}+\frac{7}{5}$

= $-\frac{5}{4}{e}^{-1}-7+\frac{5}{4}{e}^{-5}+\frac{7}{5}$

= $-\frac{5}{4}{e}^{-1}+\frac{5}{4}{e}^{-5}-7+\frac{7}{5}$

= $-\frac{5}{4}{e}^{-1}+\frac{5}{4}{e}^{-5}-\frac{28}{5}$

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

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

= $3e^{-2}-3e$

= $3e^{-2}-3e$

= ${\left[-\frac{1}{2}{x}^{-1}+3e^x\right]}_2^3$

= ${\left[-\frac{1}{2x}+3e^x\right]}_2^3$

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

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

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

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

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

= $3e^3-\frac{1}{6}-3e^2+\frac{1}{4}$

= $3e^3-3e^2-\frac{1}{6}+\frac{1}{4}$

= $3e^3-3e^2+\frac{1}{12}$

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

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

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

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

= $0+1$

= $1$