$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 6)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -9 -4.5 +2 +4 = -7.5

= ${\left[-\frac{5}{2}{x}^2-4x\right]}_{-2}^1$

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

= $-\frac{5}{2}\cdot 1-4-\left(-\frac{5}{2}\cdot 4+8\right)$

= $-\frac{5}{2}-4-\left(-10+8\right)$

= $-\frac{5}{2}-\frac{8}{2}-1·\left(-2\right)$

= $-\frac{13}{2}+2$

= $-\frac{13}{2}+\frac{4}{2}$

= $-\frac{9}{2}$

= ${\left[-2x^{-1}+\frac{5}{2}\cdot \sin \left(x\right)\right]}_{\pi }^{\frac{3}{2}\pi }$

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

$=$ $-\frac{2}{\frac{3}{2}\pi }+\frac{5}{2}\cdot \sin \left(\frac{3}{2}\pi \right)-\left(-\frac{2}{\pi }+\frac{5}{2}\cdot \sin \left(\pi \right)\right)$

= $-\frac{2}{\frac{3}{2}\pi }+\frac{5}{2}\cdot \left(-1\right)-\left(-\frac{2}{\pi }+\frac{5}{2}\cdot 0\right)$

= $-\frac{2}{\frac{3}{2}\pi }-\frac{5}{2}-\left(-\frac{2}{\pi }+0\right)$

= $-\frac{5}{2}-\frac{4}{3\pi }+\frac{2}{\pi }$

= $-\frac{5}{2}+\frac{2}{3\pi }$

= ${\left[-{\left(2x-5\right)}^{-1}\right]}_4^7$

= ${\left[-\frac{1}{2x-5}\right]}_4^7$

$=$ $-\frac{1}{2\cdot 7-5}+\frac{1}{2\cdot 4-5}$

= $-\frac{1}{14-5}+\frac{1}{8-5}$

= $-\frac{1}{9}+\frac{1}{3}$

= $-\left(\frac{1}{9}\right)+\frac{1}{3}$

= $-\frac{1}{9}+\frac{3}{9}$

= $\frac{2}{9}$

= ${\left[-3\cdot \sin \left(x\right)+\frac{1}{12}{e}^{-3x}\right]}_0^{\frac{3}{2}\pi }$

$=$ $-3\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{1}{12}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(-3\cdot \sin \left(0\right)+\frac{1}{12}{e}^{-3\cdot \left(0\right)}\right)$

= $-3\cdot \left(-1\right)+\frac{1}{12}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(-3\cdot 0+\frac{1}{12}{e}^0\right)$

= $3+\frac{1}{12}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(0+\frac{1}{12}\right)$

= $\frac{1}{12}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}+3-\left(0+\frac{1}{12}\right)$

= $\frac{1}{12}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}+3-\frac{1}{12}$

= $\frac{1}{12}{e}^{-\frac{9}{2}\pi }+\frac{35}{12}$

= ${\left[-\frac{1}{9}{\left(-3x+3\right)}^3\right]}_0^2$

$=$ $-\frac{1}{9}\cdot {\left(-3\cdot 2+3\right)}^3+\frac{1}{9}\cdot {\left(-3\cdot 0+3\right)}^3$

= $-\frac{1}{9}\cdot {\left(-6+3\right)}^3+\frac{1}{9}\cdot {\left(0+3\right)}^3$

= $-\frac{1}{9}\cdot {\left(-3\right)}^3+\frac{1}{9}\cdot 3^3$

= $-\frac{1}{9}\cdot \left(-27\right)+\frac{1}{9}\cdot 27$

= $3+3$

= $6$