$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 3 = 3 ⋅ 3 = 9.

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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-3}}{2}$ = -1.5.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 8) ⋅ ( - 1 ) = 2 ⋅ ( - 1 ) = -2.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 9 +3 -1.5 -2 = 8.5

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

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

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

= $\frac{3}{2}+1-\left(\frac{27}{2}+3\right)$

= $\frac{3}{2}+\frac{2}{2}-\left(\frac{27}{2}+\frac{6}{2}\right)$

= $\frac{5}{2}-1·\frac{33}{2}$

= $\frac{5}{2}-\frac{33}{2}$

= $-14$

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

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

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

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

= $\frac{2}{3}\cdot {\pi }^3+0$

= $\frac{2}{3}\cdot {\pi }^3$

= ${\left[-\frac{1}{3}{\left(-3x+6\right)}^3\right]}_0^2$

$=$ $-\frac{1}{3}\cdot {\left(-3\cdot 2+6\right)}^3+\frac{1}{3}\cdot {\left(-3\cdot 0+6\right)}^3$

= $-\frac{1}{3}\cdot {\left(-6+6\right)}^3+\frac{1}{3}\cdot {\left(0+6\right)}^3$

= $-\frac{1}{3}\cdot 0^3+\frac{1}{3}\cdot 6^3$

= $-\frac{1}{3}\cdot 0+\frac{1}{3}\cdot 216$

= $0+72$

= $72$

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

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

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

= $-\frac{1}{\pi }-\left(-1\right)-\left(-\frac{1}{\frac{1}{2}\pi }-0\right)$

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

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

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

= ${\left[\frac{2}{3}{\left(-3x+5\right)}^{-1}\right]}_2^5$

= ${\left[\frac{2}{3\left(-3x+5\right)}\right]}_2^5$

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

= $\frac{2}{3\left(-15+5\right)}-\frac{2}{3\left(-6+5\right)}$

= $\frac{2}{3\left(-10\right)}-\frac{2}{3\left(-1\right)}$

= $\frac{2}{3}\cdot \left(-\frac{1}{10}\right)-\frac{2}{3}\cdot \left(-1\right)$

= $-\frac{1}{15}+\frac{2}{3}$

= $-\frac{1}{15}+\frac{10}{15}$

= $\frac{3}{5}$

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

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

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

= $-128+32t^2-\left(0+0\right)$

= $32t^2-128+0$

= $32t^2-128$

Dieses Integral $32t^2-128$ muss nun gleich $264$ sein:

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