$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ 2 = 3 ⋅ 2 = 6.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 6 +2 -1 = 7

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

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

= $\frac{5}{2}\cdot 4+6-\left(\frac{5}{2}\cdot 1-3\right)$

= $10+6-\left(\frac{5}{2}-3\right)$

= $16-\left(\frac{5}{2}-\frac{6}{2}\right)$

= $16-1·\left(-\frac{1}{2}\right)$

= $16+\frac{1}{2}$

= $\frac{32}{2}+\frac{1}{2}$

= $\frac{33}{2}$

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

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

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

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

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

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

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

= $\frac{1}{2}{e}^{2\pi }-\frac{9}{2}$

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

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

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

= $0-1$

= $-1$

= ${\left[-\frac{1}{4}{x}^6+\frac{15}{8}{x}^{\frac{4}{3}}\right]}_4^9$

= ${\left[-\frac{1}{4}{x}^6+\frac{15}{8}{\left(\sqrt[3]{x}\right)}^4\right]}_4^9$

$=$ $-\frac{1}{4}\cdot 9^6+\frac{15}{8}{\left(\sqrt[3]{9}\right)}^4-\left(-\frac{1}{4}\cdot 4^6+\frac{15}{8}{\left(\sqrt[3]{4}\right)}^4\right)$

= $-\frac{1}{4}\cdot 531441+\frac{15}{8}\cdot 1^4-\left(-\frac{1}{4}\cdot 4096+\frac{15}{8}\cdot 1^4\right)$

= $-\frac{531441}{4}+\frac{15}{8}\cdot 1-\left(-1024+\frac{15}{8}\cdot 1\right)$

= $-\frac{531441}{4}+\frac{15}{8}-\left(-1024+\frac{15}{8}\right)$

= $-\frac{1062882}{8}+\frac{15}{8}-\left(-\frac{8192}{8}+\frac{15}{8}\right)$

= $-\frac{1062867}{8}-1·\left(-\frac{8177}{8}\right)$

= $-\frac{1062867}{8}+\frac{8177}{8}$

= $-\frac{527345}{4}$

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

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

= $\frac{1}{2}\cdot \sin \left(\frac{7}{2}\pi \right)-\frac{1}{2}\cdot \sin \left(\frac{3}{2}\pi \right)$

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

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

= 0