$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 4 = 2 ⋅ 4 = 8.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{4\; +\; <mn>5</mn>}}{2}$ = 2 ⋅ 4.5 = 9.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₅ = (10 - 8) ⋅ 5 = 2 ⋅ 5 = 10.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ + I₅ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 +4 +8 +9 +10 = 28

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

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

= $2\cdot 4+8-\left(2\cdot 1+4\right)$

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

= $16-1·6$

= $16-6$

= $10$

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

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

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

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

= $3+\frac{3}{2}\cdot {\pi }^3-\frac{3}{16}\cdot {\pi }^3$

= $3+\frac{21}{16}\cdot {\pi }^3$

= ${\left[-\frac{2}{3}{\left(-x+3\right)}^{\frac{3}{2}}\right]}_{-22}^{-13}$

= ${\left[-\frac{2}{3}{\left(\sqrt{-x+3}\right)}^3\right]}_{-22}^{-13}$

$=$ $-\frac{2}{3}{\left(\sqrt{-\left(-13\right)+3}\right)}^3+\frac{2}{3}{\left(\sqrt{-\left(-22\right)+3}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{13+3}\right)}^3+\frac{2}{3}{\left(\sqrt{22+3}\right)}^3$

= $-\frac{2}{3}{\left(\sqrt{16}\right)}^3+\frac{2}{3}{\left(\sqrt{25}\right)}^3$

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

= $-\frac{2}{3}\cdot 64+\frac{2}{3}\cdot 125$

= $-\frac{128}{3}+\frac{250}{3}$

= $\frac{122}{3}$

= ${\left[-\frac{9}{2}\cdot \sin \left(x\right)+\frac{7}{8}{e}^{2x}\right]}_0^{\frac{1}{2}\pi }$

$=$ $-\frac{9}{2}\cdot \sin \left(\frac{1}{2}\pi \right)+\frac{7}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}-\left(-\frac{9}{2}\cdot \sin \left(0\right)+\frac{7}{8}{e}^{2\cdot \left(0\right)}\right)$

= $-\frac{9}{2}\cdot 1+\frac{7}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}-\left(-\frac{9}{2}\cdot 0+\frac{7}{8}{e}^0\right)$

= $-\frac{9}{2}+\frac{7}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}-\left(0+\frac{7}{8}\right)$

= $\frac{7}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}-\frac{9}{2}-\left(0+\frac{7}{8}\right)$

= $\frac{7}{8}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}-\frac{9}{2}-\frac{7}{8}$

= $\frac{7}{8}{e}^{\pi }-\frac{43}{8}$

= ${\left[-\frac{2}{3}{e}^{3x-5}\right]}_2^4$

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

= $-\frac{2}{3}{e}^{12-5}+\frac{2}{3}{e}^{6-5}$

= $-\frac{2}{3}{e}^7+\frac{2}{3}e$

= $-\frac{2}{3}{e}^7+\frac{2}{3}e$