$$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

Somit gilt:

$$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ = -3 = -3

= ${\left[2x^2+x\right]}_{-2}^1$

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

= $2\cdot 1+1-\left(2\cdot 4-2\right)$

= $2+1-\left(8-2\right)$

= $3-1·6$

= $3-6$

= $-3$

= ${\left[-2\cdot \sin \left(x\right)-\frac{18}{5}{x}^{\frac{5}{4}}\right]}_4^{16}$

= ${\left[-2\cdot \sin \left(x\right)-\frac{18}{5}{\left(\sqrt[4]{x}\right)}^5\right]}_4^{16}$

$=$ $-2\cdot \sin \left(16\right)-\frac{18}{5}{\left(\sqrt[4]{16}\right)}^5-\left(-2\cdot \sin \left(4\right)-\frac{18}{5}{\left(\sqrt[4]{4}\right)}^5\right)$

= $-2\cdot \sin \left(16\right)-\frac{18}{5}\cdot 2^5-\left(-2\cdot \sin \left(4\right)-\frac{18}{5}\cdot 1^5\right)$

= $-2\cdot \sin \left(16\right)-\frac{18}{5}\cdot 32-\left(-2\cdot \sin \left(4\right)-\frac{18}{5}\cdot 1\right)$

= $-2\cdot \sin \left(16\right)-\frac{576}{5}-\left(-2\cdot \sin \left(4\right)-\frac{18}{5}\right)$

= $-2\cdot \sin \left(16\right)-\frac{576}{5}-1·\left(-2\cdot \sin \left(4\right)\right)-1·\left(-\frac{18}{5}\right)$

= $-2\cdot \sin \left(16\right)-\frac{576}{5}+2\cdot \sin \left(4\right)+\frac{18}{5}$

= $-2\cdot \sin \left(16\right)+2\cdot \sin \left(4\right)-\frac{576}{5}+\frac{18}{5}$

= $-2\cdot \sin \left(16\right)+2\cdot \sin \left(4\right)-\frac{558}{5}$

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

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

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

= $-\left(-1\right)+1$

= $1+1$

= $2$

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

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

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

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

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

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

= $\frac{1}{2}-\frac{1}{4{\pi }^2}-1·\left(-\frac{1}{2}\right)-1·\left(-\frac{1}{{\pi }^2}\right)$

= $\frac{1}{2}-\frac{1}{4{\pi }^2}+\frac{1}{2}+\frac{1}{{\pi }^2}$

= $\frac{1}{2}+\frac{1}{2}-\frac{1}{4{\pi }^2}+\frac{1}{{\pi }^2}$

= $1+\frac{3}{4{\pi }^2}$

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

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

= $\frac{1}{2}\cdot \cos \left(-\frac{9}{2}\pi \right)-\frac{1}{2}\cdot \cos \left(-\frac{5}{2}\pi \right)$

= $\frac{1}{2}\cdot 0-\frac{1}{2}\cdot 0$

= $0+0$

= 0

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

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

= $-\frac{1}{4}\cdot 256+\frac{1}{2}{t}^2\cdot 16-\left(-\frac{1}{4}\cdot 0+\frac{1}{2}{t}^2\cdot 0\right)$

= $-64+8t^2-\left(0+0\right)$

= $8t^2-64+0$

= $8t^2-64$

Dieses Integral $8t^2-64$ muss nun gleich $-32$ sein:

Der gesuchte t-Wert ist somit $2$.