$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-2}}{2}$ = -1.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

I₄ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 6) ⋅ $\frac{\mathrm{-1\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 2 ) = -6.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -1 -2 -6 = -9

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

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

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

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

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

= $\frac{7}{2}+4$

= $\frac{7}{2}+\frac{8}{2}$

= $\frac{15}{2}$

= ${\left[\sin \left(x\right)+\frac{7}{4}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\sin \left(\frac{3}{2}\pi \right)+\frac{7}{4}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\sin \left(\frac{1}{2}\pi \right)+\frac{7}{4}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-1+\frac{7}{4}\cdot 0-\left(1+\frac{7}{4}\cdot 0\right)$

= $-1+0-\left(1+0\right)$

= $-1-1$

= $-2$

= ${\left[\frac{3}{4}{\left(-x+2\right)}^4+3x^2\right]}_0^1$

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

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

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

= $\frac{3}{4}+3-\left(\frac{3}{4}\cdot 16+0\right)$

= $\frac{3}{4}+\frac{12}{4}-\left(12+0\right)$

= $\frac{15}{4}-12$

= $\frac{15}{4}-\frac{48}{4}$

= $-\frac{33}{4}$

= ${\left[-\frac{3}{10}{x}^5-\frac{5}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{3}{10}\cdot {\left(\frac{3}{2}\pi \right)}^5-\frac{5}{3}\cdot \cos \left(\frac{3}{2}\pi \right)-\left(-\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5-\frac{5}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{3}{10}\cdot {\left(\frac{3}{2}\pi \right)}^5-\frac{5}{3}\cdot 0-\left(-\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5-\frac{5}{3}\cdot 0\right)$

= $-\frac{3}{10}\cdot {\left(\frac{3}{2}\pi \right)}^5+0-\left(-\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5+0\right)$

= $-\frac{729}{320}\cdot {\pi }^5+\frac{3}{320}\cdot {\pi }^5$

= $-\frac{363}{160}\cdot {\pi }^5$

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

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

= $-\frac{1}{4}\cdot {\left(12-5\right)}^4+\frac{1}{4}\cdot {\left(3-5\right)}^4$

= $-\frac{1}{4}\cdot 7^4+\frac{1}{4}\cdot {\left(-2\right)}^4$

= $-\frac{1}{4}\cdot 2401+\frac{1}{4}\cdot 16$

= $-\frac{2401}{4}+4$

= $-\frac{2401}{4}+\frac{16}{4}$

= $-\frac{2385}{4}$