$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

I₂ = $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(6\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(9\; -\; 6)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{6}{2}$ = 3.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -12 -6 +3 = -15

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

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

= $\frac{5}{3}\cdot 8+8-\left(\frac{5}{3}\cdot 0+0\right)$

= $\frac{40}{3}+8-\left(0+0\right)$

= $\frac{40}{3}+\frac{24}{3}+0$

= $\frac{64}{3}$

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

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

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

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

= $-\frac{81}{64}\cdot {\pi }^4+\frac{1}{64}\cdot {\pi }^4$

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

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

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

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

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

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

= $1$

= ${\left[4\cdot \cos \left(x\right)+\frac{5}{12}{e}^{3x}\right]}_0^{\frac{3}{2}\pi }$

$=$ $4\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{5}{12}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(4\cdot \cos \left(0\right)+\frac{5}{12}{e}^{3\cdot \left(0\right)}\right)$

= $4\cdot 0+\frac{5}{12}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(4\cdot 1+\frac{5}{12}{e}^0\right)$

= $0+\frac{5}{12}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(4+\frac{5}{12}\right)$

= $\frac{5}{12}{e}^{\frac{9}{2}\pi }-\left(\frac{48}{12}+\frac{5}{12}\right)$

= $\frac{5}{12}{e}^{\frac{9}{2}\pi }-1·\frac{53}{12}$

= $\frac{5}{12}{e}^{\frac{9}{2}\pi }-\frac{53}{12}$

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

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

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

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

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

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

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

= $\frac{28}{3}$

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

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

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

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

= $8t^2-64+0$

= $8t^2-64$

Dieses Integral $8t^2-64$ muss nun gleich $34$ sein:

Der gesuchte t-Wert ist somit $\frac{7}{2}$ = 3,5.