$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 3.5 ) = -10.5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 3 -4 -8 -10.5 = -19.5

= ${\left[-\frac{4}{3}{x}^3-2x\right]}_{-3}^0$

$=$ $-\frac{4}{3}\cdot 0^3-2\cdot 0-\left(-\frac{4}{3}\cdot {\left(-3\right)}^3-2\cdot \left(-3\right)\right)$

= $-\frac{4}{3}\cdot 0+0-\left(-\frac{4}{3}\cdot \left(-27\right)+6\right)$

= $0+0-\left(36+6\right)$

= $0-1·42$

= $-42$

= ${\left[\frac{5}{6}{x}^6+9\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $\frac{5}{6}\cdot {\pi }^6+9\cdot \sin \left(\pi \right)-\left(\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+9\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $\frac{5}{6}\cdot {\pi }^6+9\cdot 0-\left(\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+9\cdot 1\right)$

= $\frac{5}{6}\cdot {\pi }^6+0-\left(\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+9\right)$

= $\frac{5}{6}\cdot {\pi }^6-\left(9+\frac{5}{384}\cdot {\pi }^6\right)$

= $-1·9-1·\frac{5}{384}\cdot {\pi }^6+\frac{5}{6}\cdot {\pi }^6$

= $-9-\frac{5}{384}\cdot {\pi }^6+\frac{5}{6}\cdot {\pi }^6$

= $-9+\frac{105}{128}\cdot {\pi }^6$

= ${\left[-\frac{1}{6}{\left(3x-5\right)}^4\right]}_1^2$

$=$ $-\frac{1}{6}\cdot {\left(3\cdot 2-5\right)}^4+\frac{1}{6}\cdot {\left(3\cdot 1-5\right)}^4$

= $-\frac{1}{6}\cdot {\left(6-5\right)}^4+\frac{1}{6}\cdot {\left(3-5\right)}^4$

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

= $-\frac{1}{6}\cdot 1+\frac{1}{6}\cdot 16$

= $-\frac{1}{6}+\frac{8}{3}$

= $-\frac{1}{6}+\frac{16}{6}$

= $\frac{5}{2}$

= ${\left[-\frac{5}{6}{x}^6+6\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{6}\cdot {\pi }^6+6\cdot \sin \left(\pi \right)-\left(-\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+6\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{6}\cdot {\pi }^6+6\cdot 0-\left(-\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+6\cdot 1\right)$

= $-\frac{5}{6}\cdot {\pi }^6+0-\left(-\frac{5}{6}\cdot {\left(\frac{1}{2}\pi \right)}^6+6\right)$

= $-\frac{5}{6}\cdot {\pi }^6-\left(6-\frac{5}{384}\cdot {\pi }^6\right)$

= $-1·6-1·\left(-\frac{5}{384}\cdot {\pi }^6\right)-\frac{5}{6}\cdot {\pi }^6$

= $-6+\frac{5}{384}\cdot {\pi }^6-\frac{5}{6}\cdot {\pi }^6$

= $-6-\frac{105}{128}\cdot {\pi }^6$

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

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

= $\frac{2}{3}\cdot \sin \left(-\frac{9}{2}\pi \right)-\frac{2}{3}\cdot \sin \left(-\frac{3}{2}\pi \right)$

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

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

= $-\frac{4}{3}$