$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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:

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

Somit gilt:

$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -6 -8 = -14

= ${\left[-\frac{1}{2}{x}^2+4x\right]}_{-3}^{-2}$

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

= $-\frac{1}{2}\cdot 4-8-\left(-\frac{1}{2}\cdot 9-12\right)$

= $-2-8-\left(-\frac{9}{2}-12\right)$

= $-10-\left(-\frac{9}{2}-\frac{24}{2}\right)$

= $-10-1·\left(-\frac{33}{2}\right)$

= $-10+\frac{33}{2}$

= $-\frac{20}{2}+\frac{33}{2}$

= $\frac{13}{2}$

= ${\left[8e^{-x}-\frac{1}{2}{x}^{-2}\right]}_1^5$

= ${\left[8e^{-x}-\frac{1}{2x^2}\right]}_1^5$

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

= $8e^{-5}-\frac{1}{2}\cdot \left(\frac{1}{25}\right)-\left(8e^{-1}-\frac{1}{2}\cdot 1\right)$

= $8e^{-5}-\frac{1}{50}-\left(8e^{-1}-\frac{1}{2}\right)$

= $-8e^{-1}-1·\left(-\frac{1}{2}\right)+8e^{-5}-\frac{1}{50}$

= $-8e^{-1}+\frac{1}{2}+8e^{-5}-\frac{1}{50}$

= $-8e^{-1}+8e^{-5}+\frac{1}{2}-\frac{1}{50}$

= $-8e^{-1}+8e^{-5}+\frac{12}{25}$

= ${\left[\frac{1}{2}{\left(-2x+4\right)}^3\right]}_2^3$

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

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

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

= $\frac{1}{2}\cdot \left(-8\right)-\frac{1}{2}\cdot 0$

= $-4+0$

= $-4$

= ${\left[-\frac{1}{3}{e}^{3x}+\frac{3}{2}{x}^2\right]}_1^2$

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

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

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

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

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

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

= $-\frac{1}{3}{e}^6+\frac{1}{3}{e}^3+\frac{9}{2}$

= ${\left[-\frac{4}{9}{\left(-3x+5\right)}^{\frac{3}{2}}\right]}_{-\frac{20}{3}}^{-\frac{11}{3}}$

= ${\left[-\frac{4}{9}{\left(\sqrt{-3x+5}\right)}^3\right]}_{-\frac{20}{3}}^{-\frac{11}{3}}$

$=$ $-\frac{4}{9}{\left(\sqrt{-3\cdot \left(-\frac{11}{3}\right)+5}\right)}^3+\frac{4}{9}{\left(\sqrt{-3\cdot \left(-\frac{20}{3}\right)+5}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{11+5}\right)}^3+\frac{4}{9}{\left(\sqrt{20+5}\right)}^3$

= $-\frac{4}{9}{\left(\sqrt{16}\right)}^3+\frac{4}{9}{\left(\sqrt{25}\right)}^3$

= $-\frac{4}{9}\cdot 4^3+\frac{4}{9}\cdot 5^3$

= $-\frac{4}{9}\cdot 64+\frac{4}{9}\cdot 125$

= $-\frac{256}{9}+\frac{500}{9}$

= $\frac{244}{9}$