$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

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

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-3\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2 ) = -4.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4 -3 -9 -4 = -12

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

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

= $-\frac{3}{2}\cdot 25-15-\left(-\frac{3}{2}\cdot 1-3\right)$

= $-\frac{75}{2}-15-\left(-\frac{3}{2}-3\right)$

= $-\frac{75}{2}-\frac{30}{2}-\left(-\frac{3}{2}-\frac{6}{2}\right)$

= $-\frac{105}{2}-1·\left(-\frac{9}{2}\right)$

= $-\frac{105}{2}+\frac{9}{2}$

= $-48$

= ${\left[\frac{8}{15}{x}^5+\frac{1}{3}{e}^{-3x}\right]}_{-1}^2$

$=$ $\frac{8}{15}\cdot 2^5+\frac{1}{3}{e}^{-3\cdot 2}-\left(\frac{8}{15}\cdot {\left(-1\right)}^5+\frac{1}{3}{e}^{-3\cdot \left(-1\right)}\right)$

= $\frac{8}{15}\cdot 32+\frac{1}{3}{e}^{-6}-\left(\frac{8}{15}\cdot \left(-1\right)+\frac{1}{3}{e}^3\right)$

= $\frac{256}{15}+\frac{1}{3}{e}^{-6}-\left(-\frac{8}{15}+\frac{1}{3}{e}^3\right)$

= $\frac{1}{3}{e}^{-6}+\frac{256}{15}-\left(\frac{1}{3}{e}^3-\frac{8}{15}\right)$

= $-\frac{1}{3}{e}^3-1·\left(-\frac{8}{15}\right)+\frac{1}{3}{e}^{-6}+\frac{256}{15}$

= $-\frac{1}{3}{e}^3+\frac{8}{15}+\frac{1}{3}{e}^{-6}+\frac{256}{15}$

= $-\frac{1}{3}{e}^3+\frac{1}{3}{e}^{-6}+\frac{8}{15}+\frac{256}{15}$

= $-\frac{1}{3}{e}^3+\frac{1}{3}{e}^{-6}+\frac{88}{5}$

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

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

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

= $-\frac{1}{3}\cdot 1+\frac{1}{3}\cdot 0$

= $-\frac{1}{3}+0$

= $-\frac{1}{3}+0$

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

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

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

= $-\frac{2}{3}{e}^3-\frac{1}{4}\cdot 1-\left(-\frac{2}{3}{e}^{-6}-\frac{1}{4}\cdot 16\right)$

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

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

= $-\frac{2}{3}{e}^3-\frac{1}{4}+\frac{2}{3}{e}^{-6}+4$

= $-\frac{2}{3}{e}^3+\frac{2}{3}{e}^{-6}-\frac{1}{4}+4$

= $-\frac{2}{3}{e}^3+\frac{2}{3}{e}^{-6}+\frac{15}{4}$

= ${\left[\frac{2}{3}{\left(2x-2\right)}^{\frac{3}{2}}\right]}_3^{\frac{27}{2}}$

= ${\left[\frac{2}{3}{\left(\sqrt{2x-2}\right)}^3\right]}_3^{\frac{27}{2}}$

$=$ $\frac{2}{3}{\left(\sqrt{2\cdot \left(\frac{27}{2}\right)-2}\right)}^3-\frac{2}{3}{\left(\sqrt{2\cdot 3-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{27-2}\right)}^3-\frac{2}{3}{\left(\sqrt{6-2}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{25}\right)}^3-\frac{2}{3}{\left(\sqrt{4}\right)}^3$

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

= $\frac{2}{3}\cdot 125-\frac{2}{3}\cdot 8$

= $\frac{250}{3}-\frac{16}{3}$

= $78$