$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (9 - 7) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4 +2 -1.5 -2 = 2.5

= ${\left[\frac{3}{2}{x}^2-5x\right]}_{-1}^1$

$=$ $\frac{3}{2}\cdot 1^2-5\cdot 1-\left(\frac{3}{2}\cdot {\left(-1\right)}^2-5\cdot \left(-1\right)\right)$

= $\frac{3}{2}\cdot 1-5-\left(\frac{3}{2}\cdot 1+5\right)$

= $\frac{3}{2}-5-\left(\frac{3}{2}+5\right)$

= $\frac{3}{2}-\frac{10}{2}-\left(\frac{3}{2}+\frac{10}{2}\right)$

= $-\frac{7}{2}-1·\frac{13}{2}$

= $-\frac{7}{2}-\frac{13}{2}$

= $-10$

= ${\left[\frac{3}{10}{x}^5+\cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\frac{3}{10}\cdot {\left(\frac{3}{2}\pi \right)}^5+\cos \left(\frac{3}{2}\pi \right)-\left(\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5+\cos \left(\frac{1}{2}\pi \right)\right)$

= $\frac{3}{10}\cdot {\left(\frac{3}{2}\pi \right)}^5+0-\left(\frac{3}{10}\cdot {\left(\frac{1}{2}\pi \right)}^5+0\right)$

= $\frac{729}{320}\cdot {\pi }^5-\frac{3}{320}\cdot {\pi }^5$

= $\frac{363}{160}\cdot {\pi }^5$

= ${\left[-\frac{3}{2}{e}^{2x-4}\right]}_2^4$

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

= $-\frac{3}{2}{e}^{8-4}+\frac{3}{2}{e}^{4-4}$

= $-\frac{3}{2}{e}^4+\frac{3}{2}{e}^0$

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

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

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

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

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

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

= $-\frac{5}{4}+\frac{2}{3}\cdot {\pi }^3-\frac{5}{4}$

= $-\frac{5}{4}-\frac{5}{4}+\frac{2}{3}\cdot {\pi }^3$

= $-\frac{5}{2}+\frac{2}{3}\cdot {\pi }^3$

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

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

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

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

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

= $1$

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

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

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

= $-8+4t^2-\left(0+0\right)$

= $4t^2-8+0$

= $4t^2-8$

Dieses Integral $4t^2-8$ muss nun gleich $-7$ sein:

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