$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

I₃ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 6) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = 3 +4 = 7

= ${\left[-\frac{5}{3}{x}^3-3x\right]}_1^5$

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

= $-\frac{5}{3}\cdot 125-15-\left(-\frac{5}{3}\cdot 1-3\right)$

= $-\frac{625}{3}-15-\left(-\frac{5}{3}-3\right)$

= $-\frac{625}{3}-\frac{45}{3}-\left(-\frac{5}{3}-\frac{9}{3}\right)$

= $-\frac{670}{3}-1·\left(-\frac{14}{3}\right)$

= $-\frac{670}{3}+\frac{14}{3}$

= $-\frac{656}{3}$

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

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

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

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

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

= $3-\frac{1}{12{\pi }^3}-\left(-3-\frac{2}{3{\pi }^3}\right)$

= $3-\frac{1}{12{\pi }^3}-1·\left(-3\right)-1·\left(-\frac{2}{3{\pi }^3}\right)$

= $3-\frac{1}{12{\pi }^3}+3+\frac{2}{3{\pi }^3}$

= $3+3-\frac{1}{12{\pi }^3}+\frac{2}{3{\pi }^3}$

= $6+\frac{7}{12{\pi }^3}$

= ${\left[\frac{2}{9}{\left(3x-4\right)}^3-\frac{5}{2}{x}^2\right]}_0^2$

$=$ $\frac{2}{9}\cdot {\left(3\cdot 2-4\right)}^3-\frac{5}{2}\cdot 2^2-\left(\frac{2}{9}\cdot {\left(3\cdot 0-4\right)}^3-\frac{5}{2}\cdot 0^2\right)$

= $\frac{2}{9}\cdot {\left(6-4\right)}^3-\frac{5}{2}\cdot 4-\left(\frac{2}{9}\cdot {\left(0-4\right)}^3-\frac{5}{2}\cdot 0\right)$

= $\frac{2}{9}\cdot 2^3-10-\left(\frac{2}{9}\cdot {\left(-4\right)}^3+0\right)$

= $\frac{2}{9}\cdot 8-10-\left(\frac{2}{9}\cdot \left(-64\right)+0\right)$

= $\frac{16}{9}-10-\left(-\frac{128}{9}+0\right)$

= $\frac{16}{9}-\frac{90}{9}-\left(-\frac{128}{9}+0\right)$

= $-\frac{74}{9}+\frac{128}{9}$

= $6$

= ${\left[3\cdot \sin \left(x\right)-4\cdot \cos \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

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

= $3\cdot 1-4\cdot 0-\left(3\cdot 0-4\cdot 1\right)$

= $3+0-\left(0-4\right)$

= $3+4$

= $7$

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

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

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

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

= $\frac{1}{3}\cdot \left(-27\right)-\frac{1}{3}\cdot 1$

= $-9-\frac{1}{3}$

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

= $-\frac{28}{3}$

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

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

= $-\frac{2}{3}\cdot 27+t\cdot 9-\left(-\frac{2}{3}\cdot 0+t\cdot 0\right)$

= $-18+9t-\left(0+0\right)$

= $9t-18+0$

= $9t-18$

Dieses Integral $9t-18$ muss nun gleich $-\frac{27}{2}$ sein:

Der gesuchte t-Wert ist somit $\frac{1}{2}$ = 0,5.