$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

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 = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 4.5 -3 -6 = -4.5

= ${\left[-2x^2+3x\right]}_0^2$

$=$ $-2\cdot 2^2+3\cdot 2-\left(-2\cdot 0^2+3\cdot 0\right)$

= $-2\cdot 4+6-\left(-2\cdot 0+0\right)$

= $-8+6-\left(0+0\right)$

= $-2+0$

= $-2$

= ${\left[-\frac{4}{5}{x}^{\frac{5}{2}}+x^{-2}\right]}_4^{16}$

= ${\left[-\frac{4}{5}{\left(\sqrt{x}\right)}^5+\frac{1}{x^2}\right]}_4^{16}$

$=$ $-\frac{4}{5}{\left(\sqrt{16}\right)}^5+\frac{1}{{16}^2}-\left(-\frac{4}{5}{\left(\sqrt{4}\right)}^5+\frac{1}{4^2}\right)$

= $-\frac{4}{5}\cdot 4^5+\frac{1}{256}-\left(-\frac{4}{5}\cdot 2^5+\frac{1}{16}\right)$

= $-\frac{4}{5}\cdot 1024+\frac{1}{256}-\left(-\frac{4}{5}\cdot 32+\frac{1}{16}\right)$

= $-\frac{4096}{5}+\frac{1}{256}-\left(-\frac{128}{5}+\frac{1}{16}\right)$

= $-\frac{1048576}{1280}+\frac{5}{1280}-\left(-\frac{2048}{80}+\frac{5}{80}\right)$

= $-\frac{1048571}{1280}-1·\left(-\frac{2043}{80}\right)$

= $-\frac{1048571}{1280}+\frac{2043}{80}$

= $-\frac{1048571}{1280}+\frac{32688}{1280}$

= $-\frac{1048571}{1280}+\frac{2043}{80}$

= $-\frac{1048571}{1280}+\frac{32688}{1280}$

= $-\frac{1015883}{1280}$

= ${\left[-\frac{1}{3}{\left(3x-5\right)}^3-\frac{3}{2}{x}^2\right]}_0^3$

$=$ $-\frac{1}{3}\cdot {\left(3\cdot 3-5\right)}^3-\frac{3}{2}\cdot 3^2-\left(-\frac{1}{3}\cdot {\left(3\cdot 0-5\right)}^3-\frac{3}{2}\cdot 0^2\right)$

= $-\frac{1}{3}\cdot {\left(9-5\right)}^3-\frac{3}{2}\cdot 9-\left(-\frac{1}{3}\cdot {\left(0-5\right)}^3-\frac{3}{2}\cdot 0\right)$

= $-\frac{1}{3}\cdot 4^3-\frac{27}{2}-\left(-\frac{1}{3}\cdot {\left(-5\right)}^3+0\right)$

= $-\frac{1}{3}\cdot 64-\frac{27}{2}-\left(-\frac{1}{3}\cdot \left(-125\right)+0\right)$

= $-\frac{64}{3}-\frac{27}{2}-\left(\frac{125}{3}+0\right)$

= $-\frac{128}{6}-\frac{81}{6}-\left(\frac{125}{3}+0\right)$

= $-\frac{209}{6}-\frac{125}{3}$

= $-\frac{209}{6}-\frac{250}{6}$

= $-\frac{209}{6}-\frac{125}{3}$

= $-\frac{209}{6}-\frac{250}{6}$

= $-\frac{153}{2}$

= ${\left[-\frac{1}{5}{x}^5+\sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

= $-\frac{1}{5}\cdot {\pi }^5+0-\left(-\frac{1}{5}\cdot {\left(\frac{1}{2}\pi \right)}^5+1\right)$

= $-\frac{1}{5}\cdot {\pi }^5-\left(1-\frac{1}{160}\cdot {\pi }^5\right)$

= $-1·1-1·\left(-\frac{1}{160}\cdot {\pi }^5\right)-\frac{1}{5}\cdot {\pi }^5$

= $-1+\frac{1}{160}\cdot {\pi }^5-\frac{1}{5}\cdot {\pi }^5$

= $-1-\frac{31}{160}\cdot {\pi }^5$

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

$=$ $\frac{1}{2}{e}^{2\cdot 4-3}-\frac{1}{2}{e}^{2\cdot 1-3}$

= $\frac{1}{2}{e}^{8-3}-\frac{1}{2}{e}^{2-3}$

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