$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ gibt des orientierten (also mit Vorzeichen behafteten) Flächeninhalt zwischen dem Graph der Funktion und der x-Achse an.

Um diesen aus dem Schaubild zu berechnen unterteilen wir die Fläche in Rechtecke, Parallelogrammen und ggf. Dreiecke:

I₁ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{\mathrm{12}}{2}$ = 6.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (9 - 6) ⋅ ( - 1 ) = 3 ⋅ ( - 1 ) = -3.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 6 -1.5 -3 = 1.5

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

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

= $2\cdot 9-12-\left(2\cdot 1+4\right)$

= $18-12-\left(2+4\right)$

= $6-1·6$

= $6-6$

= 0

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

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

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

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

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

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

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

= $-\frac{9}{4}-\frac{3}{2{\pi }^2}$

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

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

= $-\frac{1}{2}{e}^{8-4}+\frac{1}{2}{e}^{2-4}$

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

= ${\left[4x^{\frac{5}{4}}+\frac{1}{3}{e}^{-3x}\right]}_1^{16}$

= ${\left[4{\left(\sqrt[4]{x}\right)}^5+\frac{1}{3}{e}^{-3x}\right]}_1^{16}$

$=$ $4{\left(\sqrt[4]{16}\right)}^5+\frac{1}{3}{e}^{-3\cdot 16}-\left(4{\left(\sqrt[4]{1}\right)}^5+\frac{1}{3}{e}^{-3\cdot 1}\right)$

= $4\cdot 2^5+\frac{1}{3}{e}^{-48}-\left(4\cdot 1^5+\frac{1}{3}{e}^{-3}\right)$

= $4\cdot 32+\frac{1}{3}{e}^{-48}-\left(4\cdot 1+\frac{1}{3}{e}^{-3}\right)$

= $128+\frac{1}{3}{e}^{-48}-\left(4+\frac{1}{3}{e}^{-3}\right)$

= $\frac{1}{3}{e}^{-48}+128-\left(\frac{1}{3}{e}^{-3}+4\right)$

= $-\frac{1}{3}{e}^{-3}+\frac{1}{3}{e}^{-48}-4+128$

= $-\frac{1}{3}{e}^{-3}+\frac{1}{3}{e}^{-48}+124$

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

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

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

= $2\cdot 0-2\cdot 1$

= $0-2$

= $-2$