$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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:

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

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

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

Somit gilt:

$$\underset{0}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -6 +2 +6 = 2

= ${\left[-\frac{4}{3}{x}^3-\frac{5}{2}{x}^2\right]}_0^4$

$=$ $-\frac{4}{3}\cdot 4^3-\frac{5}{2}\cdot 4^2-\left(-\frac{4}{3}\cdot 0^3-\frac{5}{2}\cdot 0^2\right)$

= $-\frac{4}{3}\cdot 64-\frac{5}{2}\cdot 16-\left(-\frac{4}{3}\cdot 0-\frac{5}{2}\cdot 0\right)$

= $-\frac{256}{3}-40-\left(0+0\right)$

= $-\frac{256}{3}-\frac{120}{3}+0$

= $-\frac{376}{3}$

= ${\left[\frac{1}{2}{x}^4+3x^{\frac{3}{2}}\right]}_0^{16}$

= ${\left[\frac{1}{2}{x}^4+3{\left(\sqrt{x}\right)}^3\right]}_0^{16}$

$=$ $\frac{1}{2}\cdot {16}^4+3{\left(\sqrt{16}\right)}^3-\left(\frac{1}{2}\cdot 0^4+3{\left(\sqrt{0}\right)}^3\right)$

= $\frac{1}{2}\cdot 65536+3\cdot 4^3-\left(\frac{1}{2}\cdot 0+3\cdot 0^3\right)$

= $32768+3\cdot 64-\left(0+3\cdot 0\right)$

= $32768+192-\left(0+0\right)$

= $32960+0$

= $32960$

= ${\left[{\left(-2x+3\right)}^{\frac{3}{2}}\right]}_{-11}^{-\frac{13}{2}}$

= ${\left[{\left(\sqrt{-2x+3}\right)}^3\right]}_{-11}^{-\frac{13}{2}}$

$=$ ${\left(\sqrt{-2\cdot \left(-\frac{13}{2}\right)+3}\right)}^3-{\left(\sqrt{-2\cdot \left(-11\right)+3}\right)}^3$

= ${\left(\sqrt{13+3}\right)}^3-{\left(\sqrt{22+3}\right)}^3$

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

= $4^3-5^3$

= $64-125$

= $-61$

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

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

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

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

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

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

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

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

= $\frac{9}{2}-\frac{1}{\pi }$

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

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

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

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

= $0-\frac{2}{3}$

= $0-\frac{2}{3}$

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