$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ 2 = 2 ⋅ 2 = 4.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = -4 -3 +2 +4 = -1

= ${\left[-x^2+4x\right]}_{-1}^0$

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

= $-0+0-\left(-1-4\right)$

= $-1·\left(-1\right)-1·\left(-4\right)$

= $1+4$

= $5$

= ${\left[\frac{9}{4}\cdot \sin \left(x\right)+2\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{9}{4}\cdot \sin \left(\frac{3}{2}\pi \right)+2\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\frac{9}{4}\cdot \sin \left(\frac{1}{2}\pi \right)+2\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{9}{4}\cdot \left(-1\right)+2\cdot 0-\left(\frac{9}{4}\cdot 1+2\cdot 0\right)$

= $-\frac{9}{4}+0-\left(\frac{9}{4}+0\right)$

= $-\frac{9}{4}+0-\left(\frac{9}{4}+0\right)$

= $-\frac{9}{4}-\frac{9}{4}$

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

= ${\left[\frac{3}{8}{\left(-2x+1\right)}^4\right]}_2^3$

$=$ $\frac{3}{8}\cdot {\left(-2\cdot 3+1\right)}^4-\frac{3}{8}\cdot {\left(-2\cdot 2+1\right)}^4$

= $\frac{3}{8}\cdot {\left(-6+1\right)}^4-\frac{3}{8}\cdot {\left(-4+1\right)}^4$

= $\frac{3}{8}\cdot {\left(-5\right)}^4-\frac{3}{8}\cdot {\left(-3\right)}^4$

= $\frac{3}{8}\cdot 625-\frac{3}{8}\cdot 81$

= $\frac{1875}{8}-\frac{243}{8}$

= $204$

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

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

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

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

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

= $\frac{1}{2}{e}^{-3\pi }-\frac{2}{3}-\frac{1}{2}{e}^{-\pi }-1·\frac{2}{3}$

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

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

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

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

$=$ $3e^{-4+2}-3e^{-1+2}$

= $3e^{-2}-3e^1$

= $3e^{-2}-3e$

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

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

= $-\frac{1}{4}\cdot 256+\frac{3}{2}{t}^2\cdot 16-\left(-\frac{1}{4}\cdot 0+\frac{3}{2}{t}^2\cdot 0\right)$

= $-64+24t^2-\left(0+0\right)$

= $24t^2-64+0$

= $24t^2-64$

Dieses Integral $24t^2-64$ muss nun gleich $230$ sein:

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