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

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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 4 -4.5 = -0.5

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

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

= $\frac{2}{3}\cdot 64+2\cdot 16-\left(\frac{2}{3}\cdot 1+2\cdot 1\right)$

= $\frac{128}{3}+32-\left(\frac{2}{3}+2\right)$

= $\frac{128}{3}+\frac{96}{3}-\left(\frac{2}{3}+\frac{6}{3}\right)$

= $\frac{224}{3}-1·\frac{8}{3}$

= $\frac{224}{3}-\frac{8}{3}$

= $72$

= ${\left[-\frac{1}{6}{e}^{-3x}-\frac{7}{3}{x}^{\frac{3}{2}}\right]}_0^{16}$

= ${\left[-\frac{1}{6}{e}^{-3x}-\frac{7}{3}{\left(\sqrt{x}\right)}^3\right]}_0^{16}$

$=$ $-\frac{1}{6}{e}^{-3\cdot 16}-\frac{7}{3}{\left(\sqrt{16}\right)}^3-\left(-\frac{1}{6}{e}^{-3\cdot 0}-\frac{7}{3}{\left(\sqrt{0}\right)}^3\right)$

= $-\frac{1}{6}{e}^{-48}-\frac{7}{3}\cdot 4^3-\left(-\frac{1}{6}{e}^0-\frac{7}{3}\cdot 0^3\right)$

= $-\frac{1}{6}{e}^{-48}-\frac{7}{3}\cdot 64-\left(-\frac{1}{6}-\frac{7}{3}\cdot 0\right)$

= $-\frac{1}{6}{e}^{-48}-\frac{448}{3}-\left(-\frac{1}{6}+0\right)$

= $-\frac{1}{6}{e}^{-48}-\frac{448}{3}-\left(-\frac{1}{6}+0\right)$

= $-\frac{1}{6}{e}^{-48}-\frac{448}{3}+\frac{1}{6}$

= $-\frac{1}{6}{e}^{-48}-\frac{895}{6}$

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

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

= $\cos \left(\frac{7}{2}\pi \right)-\cos \left(\frac{1}{2}\pi \right)$

= $0-0$

= 0

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

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

= $3\cdot \left(-1\right)+0-\left(3\cdot 1+0\right)$

= $-3+0-\left(3+0\right)$

= $-3-3$

= $-6$

= ${\left[\frac{1}{2}{\left(2x-1\right)}^3+2x\right]}_1^4$

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

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

= $\frac{1}{2}\cdot 7^3+8-\left(\frac{1}{2}\cdot 1^3+2\right)$

= $\frac{1}{2}\cdot 343+8-\left(\frac{1}{2}\cdot 1+2\right)$

= $\frac{343}{2}+8-\left(\frac{1}{2}+2\right)$

= $\frac{343}{2}+\frac{16}{2}-\left(\frac{1}{2}+\frac{4}{2}\right)$

= $\frac{359}{2}-1·\frac{5}{2}$

= $\frac{359}{2}-\frac{5}{2}$

= $177$