$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

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

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 3 -3 -6 = -6

= ${\left[\frac{4}{3}{x}^3+x^2\right]}_{-2}^0$

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

= $\frac{4}{3}\cdot 0+0-\left(\frac{4}{3}\cdot \left(-8\right)+4\right)$

= $0+0-\left(-\frac{32}{3}+4\right)$

= $0-\left(-\frac{32}{3}+\frac{12}{3}\right)$

= $-1·\left(-\frac{20}{3}\right)$

= $\frac{20}{3}$

= ${\left[-\frac{4}{5}{x}^{\frac{5}{4}}-2e^{2x}\right]}_0^4$

= ${\left[-\frac{4}{5}{\left(\sqrt[4]{x}\right)}^5-2e^{2x}\right]}_0^4$

$=$ $-\frac{4}{5}{\left(\sqrt[4]{4}\right)}^5-2e^{2\cdot 4}-\left(-\frac{4}{5}{\left(\sqrt[4]{0}\right)}^5-2e^{2\cdot 0}\right)$

= $-\frac{4}{5}\cdot 1^5-2e^8-\left(-\frac{4}{5}\cdot 0^5-2e^0\right)$

= $-\frac{4}{5}\cdot 1-2e^8-\left(-\frac{4}{5}\cdot 0-2\right)$

= $-\frac{4}{5}-2e^8-\left(0-2\right)$

= $-2e^8-\frac{4}{5}+2$

= $-2e^8+\frac{6}{5}$

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

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

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

= $\frac{1}{3}\cdot 7^3-\frac{1}{3}\cdot 1^3$

= $\frac{1}{3}\cdot 343-\frac{1}{3}\cdot 1$

= $\frac{343}{3}-\frac{1}{3}$

= $114$

= ${\left[-\frac{5}{3}{e}^{-3x}+5x\right]}_1^4$

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

= $-\frac{5}{3}{e}^{-12}+20-\left(-\frac{5}{3}{e}^{-3}+5\right)$

= $\frac{5}{3}{e}^{-3}-1·5-\frac{5}{3}{e}^{-12}+20$

= $\frac{5}{3}{e}^{-3}-5-\frac{5}{3}{e}^{-12}+20$

= $\frac{5}{3}{e}^{-3}-\frac{5}{3}{e}^{-12}-5+20$

= $\frac{5}{3}{e}^{-3}-\frac{5}{3}{e}^{-12}+15$

= ${\left[-\frac{1}{3}{\left(-3x+6\right)}^3+x^2\right]}_2^3$

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

= $-\frac{1}{3}\cdot {\left(-9+6\right)}^3+9-\left(-\frac{1}{3}\cdot {\left(-6+6\right)}^3+4\right)$

= $-\frac{1}{3}\cdot {\left(-3\right)}^3+9-\left(-\frac{1}{3}\cdot 0^3+4\right)$

= $-\frac{1}{3}\cdot \left(-27\right)+9-\left(-\frac{1}{3}\cdot 0+4\right)$

= $9+9-\left(0+4\right)$

= $18-4$

= $14$