$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₄ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 6) ⋅ 4 = 3 ⋅ 4 = 12.

Somit gilt:

$$\underset{2}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -4 +4 +12 = 12

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

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

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

= $-\frac{5}{2}+1-\left(-\frac{45}{2}-3\right)$

= $-\frac{5}{2}+\frac{2}{2}-\left(-\frac{45}{2}-\frac{6}{2}\right)$

= $-\frac{3}{2}-1·\left(-\frac{51}{2}\right)$

= $-\frac{3}{2}+\frac{51}{2}$

= $24$

= ${\left[-\frac{3}{2}{e}^{2x}-4x^{-1}\right]}_2^4$

= ${\left[-\frac{3}{2}{e}^{2x}-\frac{4}{x}\right]}_2^4$

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

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

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

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

= $-\frac{3}{2}{e}^8-1+\frac{3}{2}{e}^4+2$

= $-\frac{3}{2}{e}^8+\frac{3}{2}{e}^4-1+2$

= $-\frac{3}{2}{e}^8+\frac{3}{2}{e}^4+1$

= ${\left[-2{\left(x-1\right)}^{\frac{3}{2}}\right]}_{10}^{26}$

= ${\left[-2{\left(\sqrt{x-1}\right)}^3\right]}_{10}^{26}$

$=$ $-2{\left(\sqrt{26-1}\right)}^3+2{\left(\sqrt{10-1}\right)}^3$

= $-2{\left(\sqrt{25}\right)}^3+2{\left(\sqrt{9}\right)}^3$

= $-2\cdot 5^3+2\cdot 3^3$

= $-2\cdot 125+2\cdot 27$

= $-250+54$

= $-196$

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

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

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

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

= $\frac{3}{2}+\frac{3}{4}\cdot {\pi }^4-\frac{3}{64}\cdot {\pi }^4$

= $\frac{3}{2}+\frac{45}{64}\cdot {\pi }^4$

= ${\left[-\frac{3}{4}{\left(x-1\right)}^4\right]}_2^3$

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

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

= $-\frac{3}{4}\cdot 16+\frac{3}{4}\cdot 1$

= $-12+\frac{3}{4}$

= $-\frac{48}{4}+\frac{3}{4}$

= $-\frac{45}{4}$