$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 4 -2 -6 = -4

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

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

= $-\frac{1}{3}\cdot \left(-8\right)-4-\left(-\frac{1}{3}\cdot \left(-27\right)-6\right)$

= $\frac{8}{3}-4-\left(9-6\right)$

= $\frac{8}{3}-\frac{12}{3}-1·3$

= $-\frac{4}{3}-3$

= $-\frac{4}{3}-\frac{9}{3}$

= $-\frac{13}{3}$

= ${\left[-\frac{9}{2}\cdot \cos \left(x\right)+\frac{1}{4}{x}^4\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $-\frac{9}{2}\cdot \left(-1\right)+\frac{1}{4}\cdot {\pi }^4-\left(-\frac{9}{2}\cdot 0+\frac{1}{4}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $\frac{9}{2}+\frac{1}{4}\cdot {\pi }^4-\left(0+\frac{1}{4}\cdot {\left(\frac{1}{2}\pi \right)}^4\right)$

= $\frac{9}{2}+\frac{1}{4}\cdot {\pi }^4-\frac{1}{64}\cdot {\pi }^4$

= $\frac{9}{2}+\frac{15}{64}\cdot {\pi }^4$

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

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

= $\frac{2}{3}\cdot 1^3-2\cdot 9-\left(\frac{2}{3}\cdot 0^3-2\cdot 4\right)$

= $\frac{2}{3}\cdot 1-18-\left(\frac{2}{3}\cdot 0-8\right)$

= $\frac{2}{3}-18-\left(0-8\right)$

= $\frac{2}{3}-\frac{54}{3}+8$

= $-\frac{52}{3}+8$

= $-\frac{52}{3}+\frac{24}{3}$

= $-\frac{28}{3}$

= ${\left[\frac{3}{5}{x}^5-\frac{2}{3}{x}^{\frac{3}{2}}\right]}_4^9$

= ${\left[\frac{3}{5}{x}^5-\frac{2}{3}{\left(\sqrt{x}\right)}^3\right]}_4^9$

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

= $\frac{3}{5}\cdot 59049-\frac{2}{3}\cdot 3^3-\left(\frac{3}{5}\cdot 1024-\frac{2}{3}\cdot 2^3\right)$

= $\frac{177147}{5}-\frac{2}{3}\cdot 27-\left(\frac{3072}{5}-\frac{2}{3}\cdot 8\right)$

= $\frac{177147}{5}-18-\left(\frac{3072}{5}-\frac{16}{3}\right)$

= $\frac{177147}{5}-\frac{90}{5}-\left(\frac{9216}{15}-\frac{80}{15}\right)$

= $\frac{177057}{5}-1·\frac{9136}{15}$

= $\frac{177057}{5}-\frac{9136}{15}$

= $\frac{531171}{15}-\frac{9136}{15}$

= $\frac{177057}{5}-\frac{9136}{15}$

= $\frac{531171}{15}-\frac{9136}{15}$

= $\frac{104407}{3}$

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

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

= $-\frac{2}{3}\cdot \cos \left(\frac{5}{2}\pi \right)+\frac{2}{3}\cdot \cos \left(\pi \right)$

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

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

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

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