$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -2 -6 = -8

= ${\left[-x^3-5x\right]}_{-2}^0$

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

= $-0+0-\left(-\left(-8\right)+10\right)$

= $0+0-\left(8+10\right)$

= $0-1·18$

= $-18$

= ${\left[\frac{16}{5}{x}^{\frac{5}{2}}+\frac{1}{2}{x}^4\right]}_0^{16}$

= ${\left[\frac{16}{5}{\left(\sqrt{x}\right)}^5+\frac{1}{2}{x}^4\right]}_0^{16}$

$=$ $\frac{16}{5}{\left(\sqrt{16}\right)}^5+\frac{1}{2}\cdot {16}^4-\left(\frac{16}{5}{\left(\sqrt{0}\right)}^5+\frac{1}{2}\cdot 0^4\right)$

= $\frac{16}{5}\cdot 4^5+\frac{1}{2}\cdot 65536-\left(\frac{16}{5}\cdot 0^5+\frac{1}{2}\cdot 0\right)$

= $\frac{16}{5}\cdot 1024+32768-\left(\frac{16}{5}\cdot 0+0\right)$

= $\frac{16384}{5}+32768-\left(0+0\right)$

= $\frac{16384}{5}+\frac{163840}{5}+0$

= $\frac{180224}{5}$

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

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

= $\cos \left(-5\pi \right)-\cos \left(-2\pi \right)$

= $-1-1$

= $-2$

= ${\left[-\frac{5}{4}\cdot \cos \left(x\right)-\frac{1}{2}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{5}{4}\cdot \cos \left(\pi \right)-\frac{1}{2}\cdot \sin \left(\pi \right)-\left(-\frac{5}{4}\cdot \cos \left(\frac{1}{2}\pi \right)-\frac{1}{2}\cdot \sin \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{5}{4}\cdot \left(-1\right)-\frac{1}{2}\cdot 0-\left(-\frac{5}{4}\cdot 0-\frac{1}{2}\cdot 1\right)$

= $\frac{5}{4}+0-\left(0-\frac{1}{2}\right)$

= $\frac{5}{4}+0-\left(0-\mathrm{0,5}\right)$

= $\frac{5}{4}+\mathrm{0,5}$

= $\frac{5}{4}+\frac{1}{2}$

= $\frac{5}{4}+\frac{2}{4}$

= $\frac{7}{4}$

= ${\left[\frac{4}{3}{\left(-x+2\right)}^{\frac{3}{2}}\right]}_{-23}^{-14}$

= ${\left[\frac{4}{3}{\left(\sqrt{-x+2}\right)}^3\right]}_{-23}^{-14}$

$=$ $\frac{4}{3}{\left(\sqrt{-\left(-14\right)+2}\right)}^3-\frac{4}{3}{\left(\sqrt{-\left(-23\right)+2}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{14+2}\right)}^3-\frac{4}{3}{\left(\sqrt{23+2}\right)}^3$

= $\frac{4}{3}{\left(\sqrt{16}\right)}^3-\frac{4}{3}{\left(\sqrt{25}\right)}^3$

= $\frac{4}{3}\cdot 4^3-\frac{4}{3}\cdot 5^3$

= $\frac{4}{3}\cdot 64-\frac{4}{3}\cdot 125$

= $\frac{256}{3}-\frac{500}{3}$

= $-\frac{244}{3}$