$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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:

I₁ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

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

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ 2 = 3 ⋅ 2 = 6.

Somit gilt:

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

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

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

= $\frac{2}{3}\cdot 1-4-\left(\frac{2}{3}\cdot \left(-27\right)+12\right)$

= $\frac{2}{3}-4-\left(-18+12\right)$

= $\frac{2}{3}-\frac{12}{3}-1·\left(-6\right)$

= $-\frac{10}{3}+6$

= $-\frac{10}{3}+\frac{18}{3}$

= $\frac{8}{3}$

= ${\left[-\frac{7}{9}{x}^{-3}+\frac{7}{9}{x}^3\right]}_1^5$

= ${\left[-\frac{7}{9x^3}+\frac{7}{9}{x}^3\right]}_1^5$

$=$ $-\frac{7}{9\cdot 5^3}+\frac{7}{9}\cdot 5^3-\left(-\frac{7}{9\cdot 1^3}+\frac{7}{9}\cdot 1^3\right)$

= $-\frac{7}{9}\cdot \left(\frac{1}{125}\right)+\frac{7}{9}\cdot 125-\left(-\frac{7}{9}\cdot 1+\frac{7}{9}\cdot 1\right)$

= $-\frac{7}{1125}+\frac{875}{9}-\left(-\frac{7}{9}+\frac{7}{9}\right)$

= $-\frac{7}{1125}+\frac{109375}{1125}-1·0$

= $\frac{12152}{125}+0$

= $\frac{12152}{125}+0$

= $\frac{12152}{125}$

= ${\left[-\frac{2}{3}{\left(3x-4\right)}^{\frac{3}{2}}\right]}_{\frac{8}{3}}^{\frac{20}{3}}$

= ${\left[-\frac{2}{3}{\left(\sqrt{3x-4}\right)}^3\right]}_{\frac{8}{3}}^{\frac{20}{3}}$

$=$ $-\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{20}{3}\right)-4}\right)}^3+\frac{2}{3}{\left(\sqrt{3\cdot \left(\frac{8}{3}\right)-4}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{20-4}\right)}^3+\frac{2}{3}{\left(\sqrt{8-4}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{16}\right)}^3+\frac{2}{3}{\left(\sqrt{4}\right)}^3$

= $-\frac{2}{3}\cdot 4^3+\frac{2}{3}\cdot 2^3$

= $-\frac{2}{3}\cdot 64+\frac{2}{3}\cdot 8$

= $-\frac{128}{3}+\frac{16}{3}$

= $-\frac{112}{3}$

= ${\left[4\cdot \sin \left(x\right)-\frac{1}{2}{e}^{-2x}\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $4\cdot 0-\frac{1}{2}{e}^{-2\cdot \pi }-\left(4\cdot 1-\frac{1}{2}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $0-\frac{1}{2}{e}^{-2\cdot \pi }-\left(4-\frac{1}{2}{e}^{-2\cdot \left(\frac{1}{2}\pi \right)}\right)$

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

= $-\frac{1}{2}{e}^{-2\pi }+\frac{1}{2}{e}^{-\pi }-1·4$

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

= $\frac{1}{2}{e}^{-\pi }-\frac{1}{2}{e}^{-2\pi }-4$

= ${\left[\frac{1}{3}{e}^{-3x+5}\right]}_1^2$

$=$ $\frac{1}{3}{e}^{-3\cdot 2+5}-\frac{1}{3}{e}^{-3\cdot 1+5}$

= $\frac{1}{3}{e}^{-6+5}-\frac{1}{3}{e}^{-3+5}$

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