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

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 = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ 3 = 3 ⋅ 3 = 9.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{3\; +\; <mn>1</mn>}}{2}$ = 3 ⋅ 2 = 6.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 +3 +9 +6 = 15

= ${\left[\frac{5}{2}{x}^2-5x\right]}_{-3}^{-2}$

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

= $\frac{5}{2}\cdot 4+10-\left(\frac{5}{2}\cdot 9+15\right)$

= $10+10-\left(\frac{45}{2}+15\right)$

= $20-\left(\frac{45}{2}+\frac{30}{2}\right)$

= $20-1·\frac{75}{2}$

= $20-\frac{75}{2}$

= $\frac{40}{2}-\frac{75}{2}$

= $-\frac{35}{2}$

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

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

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

= $\frac{1}{9}-e^{-9}-\left(\frac{1}{4}-e^{-6}\right)$

= $\frac{1}{9}-e^{-9}-1·\frac{1}{4}+e^{-6}$

= $\frac{1}{9}-e^{-9}-\frac{1}{4}+e^{-6}$

= $e^{-6}-e^{-9}+\frac{1}{9}-\frac{1}{4}$

= $e^{-6}-e^{-9}-\frac{5}{36}$

= ${\left[-\frac{1}{6}{\left(2x-4\right)}^3\right]}_0^3$

$=$ $-\frac{1}{6}\cdot {\left(2\cdot 3-4\right)}^3+\frac{1}{6}\cdot {\left(2\cdot 0-4\right)}^3$

= $-\frac{1}{6}\cdot {\left(6-4\right)}^3+\frac{1}{6}\cdot {\left(0-4\right)}^3$

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

= $-\frac{1}{6}\cdot 8+\frac{1}{6}\cdot \left(-64\right)$

= $-\frac{4}{3}-\frac{32}{3}$

= $-12$

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

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

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

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

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

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

= $-1-\frac{21}{128}\cdot {\pi }^6$

= ${\left[-\frac{1}{3}{\left(3x-5\right)}^3\right]}_0^3$

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

= $-\frac{1}{3}\cdot {\left(9-5\right)}^3+\frac{1}{3}\cdot {\left(0-5\right)}^3$

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

= $-\frac{1}{3}\cdot 64+\frac{1}{3}\cdot \left(-125\right)$

= $-\frac{64}{3}-\frac{125}{3}$

= $-63$