$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -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 = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 2 = 2 ⋅ 2 = 4.

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 +4 = 0

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

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

= $-\frac{5}{3}\cdot \left(-1\right)+\frac{3}{2}\cdot 1-\left(-\frac{5}{3}\cdot \left(-27\right)+\frac{3}{2}\cdot 9\right)$

= $\frac{5}{3}+\frac{3}{2}-\left(45+\frac{27}{2}\right)$

= $\frac{10}{6}+\frac{9}{6}-\left(\frac{90}{2}+\frac{27}{2}\right)$

= $\frac{19}{6}-1·\frac{117}{2}$

= $\frac{19}{6}-\frac{117}{2}$

= $\frac{19}{6}-\frac{351}{6}$

= $-\frac{166}{3}$

= ${\left[\frac{1}{3}{e}^{-x}+5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

= $\frac{1}{3}{e}^{-\pi }-5-\frac{1}{3}{e}^{-\frac{1}{2}\pi }$

= ${\left[\frac{1}{4}{\left(-3x+6\right)}^4\right]}_0^3$

$=$ $\frac{1}{4}\cdot {\left(-3\cdot 3+6\right)}^4-\frac{1}{4}\cdot {\left(-3\cdot 0+6\right)}^4$

= $\frac{1}{4}\cdot {\left(-9+6\right)}^4-\frac{1}{4}\cdot {\left(0+6\right)}^4$

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

= $\frac{1}{4}\cdot 81-\frac{1}{4}\cdot 1296$

= $\frac{81}{4}-324$

= $\frac{81}{4}-\frac{1296}{4}$

= $-\frac{1215}{4}$

= ${\left[\frac{3}{8}{x}^{\frac{4}{3}}-\frac{2}{3}\cdot \sin \left(x\right)\right]}_1^4$

= ${\left[\frac{3}{8}{\left(\sqrt[3]{x}\right)}^4-\frac{2}{3}\cdot \sin \left(x\right)\right]}_1^4$

$=$ $\frac{3}{8}{\left(\sqrt[3]{4}\right)}^4-\frac{2}{3}\cdot \sin \left(4\right)-\left(\frac{3}{8}{\left(\sqrt[3]{1}\right)}^4-\frac{2}{3}\cdot \sin \left(1\right)\right)$

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

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

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

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

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

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

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

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

= ${\left[2e^{-x+2}\right]}_2^5$

$=$ $2e^{-5+2}-2e^{-2+2}$

= $2e^{-3}-2e^0$

= $2e^{-3}-2$