$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -9 -3 +1 +2 = -9

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

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

= $\left(-8\right)-\frac{3}{2}\cdot 4-\left(\left(-27\right)-\frac{3}{2}\cdot 9\right)$

= $-8-6-\left(-27-\frac{27}{2}\right)$

= $-14-\left(-\frac{54}{2}-\frac{27}{2}\right)$

= $-14-1·\left(-\frac{81}{2}\right)$

= $-14+\frac{81}{2}$

= $-\frac{28}{2}+\frac{81}{2}$

= $\frac{53}{2}$

= ${\left[3x^{\frac{4}{3}}-\frac{1}{2}{x}^{-2}\right]}_1^9$

= ${\left[3{\left(\sqrt[3]{x}\right)}^4-\frac{1}{2x^2}\right]}_1^9$

$=$ $3{\left(\sqrt[3]{9}\right)}^4-\frac{1}{2\cdot 9^2}-\left(3{\left(\sqrt[3]{1}\right)}^4-\frac{1}{2\cdot 1^2}\right)$

= $3\cdot 1^4-\frac{1}{2}\cdot \left(\frac{1}{81}\right)-\left(3\cdot 1^4-\frac{1}{2}\cdot 1\right)$

= $3\cdot 1-\frac{1}{162}-\left(3\cdot 1-\frac{1}{2}\right)$

= $3-\frac{1}{162}-\left(3-\frac{1}{2}\right)$

= $\frac{486}{162}-\frac{1}{162}-\left(\frac{6}{2}-\frac{1}{2}\right)$

= $\frac{485}{162}-1·\frac{5}{2}$

= $\frac{485}{162}-\frac{5}{2}$

= $\frac{485}{162}-\frac{405}{162}$

= $\frac{485}{162}-\frac{5}{2}$

= $\frac{485}{162}-\frac{405}{162}$

= $\frac{40}{81}$

= ${\left[\frac{1}{3}{e}^{-3x+3}\right]}_2^5$

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

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

= $\frac{1}{3}{e}^{-12}-\frac{1}{3}{e}^{-3}$

= ${\left[-\cos \left(x\right)-\frac{8}{3}{e}^{-3x}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\cos \left(\frac{3}{2}\pi \right)-\frac{8}{3}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(-\cos \left(\frac{1}{2}\pi \right)-\frac{8}{3}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-0-\frac{8}{3}{e}^{-3\cdot \left(\frac{3}{2}\pi \right)}-\left(-0-\frac{8}{3}{e}^{-3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{8}{3}{e}^{-\frac{9}{2}\pi }+\frac{8}{3}{e}^{-\frac{3}{2}\pi }$

= $\frac{8}{3}{e}^{-\frac{3}{2}\pi }-\frac{8}{3}{e}^{-\frac{9}{2}\pi }$

= ${\left[-\frac{2}{3}{\left(3x-5\right)}^{\frac{3}{2}}\right]}_7^{10}$

= ${\left[-\frac{2}{3}{\left(\sqrt{3x-5}\right)}^3\right]}_7^{10}$

$=$ $-\frac{2}{3}{\left(\sqrt{3\cdot 10-5}\right)}^3+\frac{2}{3}{\left(\sqrt{3\cdot 7-5}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{30-5}\right)}^3+\frac{2}{3}{\left(\sqrt{21-5}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{25}\right)}^3+\frac{2}{3}{\left(\sqrt{16}\right)}^3$

= $-\frac{2}{3}\cdot 5^3+\frac{2}{3}\cdot 4^3$

= $-\frac{2}{3}\cdot 125+\frac{2}{3}\cdot 64$

= $-\frac{250}{3}+\frac{128}{3}$

= $-\frac{122}{3}$