$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

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

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ 1 = 2 ⋅ 1 = 2.

Somit gilt:

$$\underset{3}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -4 +1 +2 = -1

= ${\left[\frac{3}{2}{x}^2-4x\right]}_1^4$

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

= $\frac{3}{2}\cdot 16-16-\left(\frac{3}{2}\cdot 1-4\right)$

= $24-16-\left(\frac{3}{2}-4\right)$

= $8-\left(\frac{3}{2}-\frac{8}{2}\right)$

= $8-1·\left(-\frac{5}{2}\right)$

= $8+\frac{5}{2}$

= $\frac{16}{2}+\frac{5}{2}$

= $\frac{21}{2}$

= ${\left[-\frac{9}{4}{e}^{-x}-\frac{5}{4}{x}^4\right]}_1^4$

$=$ $-\frac{9}{4}{e}^{-4}-\frac{5}{4}\cdot 4^4-\left(-\frac{9}{4}{e}^{-1}-\frac{5}{4}\cdot 1^4\right)$

= $-\frac{9}{4}{e}^{-4}-\frac{5}{4}\cdot 256-\left(-\frac{9}{4}{e}^{-1}-\frac{5}{4}\cdot 1\right)$

= $-\frac{9}{4}{e}^{-4}-320-\left(-\frac{9}{4}{e}^{-1}-\frac{5}{4}\right)$

= $\frac{9}{4}{e}^{-1}-1·\left(-\frac{5}{4}\right)-\frac{9}{4}{e}^{-4}-320$

= $\frac{9}{4}{e}^{-1}+\frac{5}{4}-\frac{9}{4}{e}^{-4}-320$

= $\frac{9}{4}{e}^{-1}-\frac{9}{4}{e}^{-4}+\frac{5}{4}-320$

= $\frac{9}{4}{e}^{-1}-\frac{9}{4}{e}^{-4}-\frac{1275}{4}$

= ${\left[-\frac{1}{9}{\left(3x-4\right)}^3\right]}_0^3$

$=$ $-\frac{1}{9}\cdot {\left(3\cdot 3-4\right)}^3+\frac{1}{9}\cdot {\left(3\cdot 0-4\right)}^3$

= $-\frac{1}{9}\cdot {\left(9-4\right)}^3+\frac{1}{9}\cdot {\left(0-4\right)}^3$

= $-\frac{1}{9}\cdot 5^3+\frac{1}{9}\cdot {\left(-4\right)}^3$

= $-\frac{1}{9}\cdot 125+\frac{1}{9}\cdot \left(-64\right)$

= $-\frac{125}{9}-\frac{64}{9}$

= $-21$

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

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

$=$ $-\frac{4}{3}{\left(\sqrt{4}\right)}^3+\sin \left(4\right)-\left(-\frac{4}{3}{\left(\sqrt{0}\right)}^3+\sin \left(0\right)\right)$

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

= $-\frac{4}{3}\cdot 8+\sin \left(4\right)-\left(-\frac{4}{3}\cdot 0+0\right)$

= $-\frac{32}{3}+\sin \left(4\right)-\left(0+0\right)$

= $\sin \left(4\right)-\frac{32}{3}+0$

= $\sin \left(4\right)-\frac{32}{3}$

= ${\left[\frac{1}{3}{\left(-2x+5\right)}^{\frac{3}{2}}\right]}_{-\frac{11}{2}}^{-2}$

= ${\left[\frac{1}{3}{\left(\sqrt{-2x+5}\right)}^3\right]}_{-\frac{11}{2}}^{-2}$

$=$ $\frac{1}{3}{\left(\sqrt{-2\cdot \left(-2\right)+5}\right)}^3-\frac{1}{3}{\left(\sqrt{-2\cdot \left(-\frac{11}{2}\right)+5}\right)}^3$

= $\frac{1}{3}{\left(\sqrt{4+5}\right)}^3-\frac{1}{3}{\left(\sqrt{11+5}\right)}^3$

= $\frac{1}{3}{\left(\sqrt{9}\right)}^3-\frac{1}{3}{\left(\sqrt{16}\right)}^3$

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

= $\frac{1}{3}\cdot 27-\frac{1}{3}\cdot 64$

= $9-\frac{64}{3}$

= $\frac{27}{3}-\frac{64}{3}$

= $-\frac{37}{3}$