$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ 4 = 2 ⋅ 4 = 8.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-4}}{2}$ = -2.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 8 +6 -2 = 12

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

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

= $-27+\frac{5}{2}\cdot 9-\left(-\left(-1\right)+\frac{5}{2}\cdot 1\right)$

= $-27+\frac{45}{2}-\left(1+\frac{5}{2}\right)$

= $-\frac{54}{2}+\frac{45}{2}-\left(\frac{2}{2}+\frac{5}{2}\right)$

= $-\frac{9}{2}-1·\frac{7}{2}$

= $-\frac{9}{2}-\frac{7}{2}$

= $-8$

= ${\left[-\frac{4}{3}\cdot \cos \left(x\right)+\frac{3}{5}{x}^5\right]}_0^{\pi }$

$=$ $-\frac{4}{3}\cdot \cos \left(\pi \right)+\frac{3}{5}\cdot {\pi }^5-\left(-\frac{4}{3}\cdot \cos \left(0\right)+\frac{3}{5}\cdot {\left(0\right)}^5\right)$

= $-\frac{4}{3}\cdot \left(-1\right)+\frac{3}{5}\cdot {\pi }^5-\left(-\frac{4}{3}\cdot 1+\frac{3}{5}\cdot 0\right)$

= $\frac{4}{3}+\frac{3}{5}\cdot {\pi }^5-\left(-\frac{4}{3}+0\right)$

= $\frac{4}{3}+\frac{3}{5}\cdot {\pi }^5-\left(-\frac{4}{3}+0\right)$

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

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

= $\frac{8}{3}+\frac{3}{5}\cdot {\pi }^5$

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

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

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

= ${\left(\sqrt{1+3}\right)}^3-{\left(\sqrt{6+3}\right)}^3$

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

= $2^3-3^3$

= $8-27$

= $-19$

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

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

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

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

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

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

= $\frac{3}{2}{e}^{3\pi }-\frac{7}{2}$

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

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

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

= $0-1$

= $-1$