$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : 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 = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 7) ⋅ 4 = 3 ⋅ 4 = 12.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 -3 +4 +12 = 9

= ${\left[\frac{4}{3}{x}^3+3x\right]}_1^4$

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

= $\frac{4}{3}\cdot 64+12-\left(\frac{4}{3}\cdot 1+3\right)$

= $\frac{256}{3}+12-\left(\frac{4}{3}+3\right)$

= $\frac{256}{3}+\frac{36}{3}-\left(\frac{4}{3}+\frac{9}{3}\right)$

= $\frac{292}{3}-1·\frac{13}{3}$

= $\frac{292}{3}-\frac{13}{3}$

= $93$

= ${\left[\frac{5}{6}{x}^{\frac{3}{2}}-\frac{9}{2}\cdot \sin \left(x\right)\right]}_1^4$

= ${\left[\frac{5}{6}{\left(\sqrt{x}\right)}^3-\frac{9}{2}\cdot \sin \left(x\right)\right]}_1^4$

$=$ $\frac{5}{6}{\left(\sqrt{4}\right)}^3-\frac{9}{2}\cdot \sin \left(4\right)-\left(\frac{5}{6}{\left(\sqrt{1}\right)}^3-\frac{9}{2}\cdot \sin \left(1\right)\right)$

= $\frac{5}{6}\cdot 2^3-\frac{9}{2}\cdot \sin \left(4\right)-\left(\frac{5}{6}\cdot 1^3-\frac{9}{2}\cdot \sin \left(1\right)\right)$

= $\frac{5}{6}\cdot 8-\frac{9}{2}\cdot \sin \left(4\right)-\left(\frac{5}{6}\cdot 1-\frac{9}{2}\cdot \sin \left(1\right)\right)$

= $\frac{20}{3}-\frac{9}{2}\cdot \sin \left(4\right)-\left(\frac{5}{6}-\frac{9}{2}\cdot \sin \left(1\right)\right)$

= $-\frac{9}{2}\cdot \sin \left(4\right)+\frac{20}{3}-\left(-\frac{9}{2}\cdot \sin \left(1\right)+\frac{5}{6}\right)$

= $-\frac{9}{2}\cdot \sin \left(4\right)+\frac{20}{3}-1·\left(-\frac{9}{2}\cdot \sin \left(1\right)\right)-1·\frac{5}{6}$

= $-\frac{9}{2}\cdot \sin \left(4\right)+\frac{20}{3}+\frac{9}{2}\cdot \sin \left(1\right)-\frac{5}{6}$

= $-\frac{9}{2}\cdot \sin \left(4\right)+\frac{9}{2}\cdot \sin \left(1\right)+\frac{20}{3}-\frac{5}{6}$

= $-\frac{9}{2}\cdot \sin \left(4\right)+\frac{9}{2}\cdot \sin \left(1\right)+\frac{35}{6}$

= ${\left[-e^{x-1}\right]}_2^3$

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

= $-e^2+e$

= $-e^2+e$

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

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

= $-4\cdot \left(-1\right)-\frac{5}{4}\cdot 0-\left(-4\cdot 0-\frac{5}{4}\cdot 1\right)$

= $4+0-\left(0-\frac{5}{4}\right)$

= $4-\left(0-\frac{5}{4}\right)$

= $4+\frac{5}{4}$

= $\frac{16}{4}+\frac{5}{4}$

= $\frac{21}{4}$

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

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

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

= $3\cdot 1-3\cdot 0$

= $3+0$

= $3$