$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

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

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4 -2 +2 +4 = 0

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

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

= $-\frac{4}{3}\cdot 1+2\cdot 1-\left(-\frac{4}{3}\cdot \left(-8\right)+2\cdot 4\right)$

= $-\frac{4}{3}+2-\left(\frac{32}{3}+8\right)$

= $-\frac{4}{3}+\frac{6}{3}-\left(\frac{32}{3}+\frac{24}{3}\right)$

= $\frac{2}{3}-1·\frac{56}{3}$

= $\frac{2}{3}-\frac{56}{3}$

= $-18$

= ${\left[x^{-2}+\frac{7}{9}{e}^{3x}\right]}_1^4$

= ${\left[\frac{1}{x^2}+\frac{7}{9}{e}^{3x}\right]}_1^4$

$=$ $\frac{1}{4^2}+\frac{7}{9}{e}^{3\cdot 4}-\left(\frac{1}{1^2}+\frac{7}{9}{e}^{3\cdot 1}\right)$

= $\frac{1}{16}+\frac{7}{9}{e}^{12}-\left(1+\frac{7}{9}{e}^3\right)$

= $\frac{1}{16}+\frac{7}{9}{e}^{12}-1·1-\frac{7}{9}{e}^3$

= $\frac{1}{16}+\frac{7}{9}{e}^{12}-1-\frac{7}{9}{e}^3$

= $\frac{7}{9}{e}^{12}-\frac{7}{9}{e}^3+\frac{1}{16}-1$

= $\frac{7}{9}{e}^{12}-\frac{7}{9}{e}^3-\frac{15}{16}$

= ${\left[\frac{4}{9}{\left(3x-3\right)}^{\frac{3}{2}}\right]}_4^{\frac{28}{3}}$

= ${\left[\frac{4}{9}{\left(\sqrt{3x-3}\right)}^3\right]}_4^{\frac{28}{3}}$

$=$ $\frac{4}{9}{\left(\sqrt{3\cdot \left(\frac{28}{3}\right)-3}\right)}^3-\frac{4}{9}{\left(\sqrt{3\cdot 4-3}\right)}^3$

= $\frac{4}{9}{\left(\sqrt{28-3}\right)}^3-\frac{4}{9}{\left(\sqrt{12-3}\right)}^3$

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

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

= $\frac{4}{9}\cdot 125-\frac{4}{9}\cdot 27$

= $\frac{500}{9}-12$

= $\frac{500}{9}-\frac{108}{9}$

= $\frac{392}{9}$

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

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

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

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

= $-\frac{1}{3}+0-2$

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

= $-\frac{1}{3}-\frac{6}{3}$

= $-\frac{7}{3}$

= ${\left[\frac{1}{9}{\left(-3x+5\right)}^3-3x^2\right]}_2^4$

$=$ $\frac{1}{9}\cdot {\left(-3\cdot 4+5\right)}^3-3\cdot 4^2-\left(\frac{1}{9}\cdot {\left(-3\cdot 2+5\right)}^3-3\cdot 2^2\right)$

= $\frac{1}{9}\cdot {\left(-12+5\right)}^3-3\cdot 16-\left(\frac{1}{9}\cdot {\left(-6+5\right)}^3-3\cdot 4\right)$

= $\frac{1}{9}\cdot {\left(-7\right)}^3-48-\left(\frac{1}{9}\cdot {\left(-1\right)}^3-12\right)$

= $\frac{1}{9}\cdot \left(-343\right)-48-\left(\frac{1}{9}\cdot \left(-1\right)-12\right)$

= $-\frac{343}{9}-48-\left(-\frac{1}{9}-12\right)$

= $-\frac{343}{9}-\frac{432}{9}-\left(-\frac{1}{9}-\frac{108}{9}\right)$

= $-\frac{775}{9}-1·\left(-\frac{109}{9}\right)$

= $-\frac{775}{9}+\frac{109}{9}$

= $-74$