$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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) ⋅ 3 = 2 ⋅ 3 = 6.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{9}{2}$ = 4.5.

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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 6 +4.5 -1.5 -2 = 7

= ${\left[x^2-x\right]}_0^1$

$=$ $1^2-1-\left(0^2-0\right)$

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

= $1-1$

= 0

= ${\left[-\frac{5}{3}{x}^3-\frac{1}{4}\cdot \sin \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

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

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

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

= $-\frac{5}{3}\cdot {\pi }^3-\left(-\frac{1}{4}-\frac{5}{24}\cdot {\pi }^3\right)$

= $-1·\left(-\frac{1}{4}\right)-1·\left(-\frac{5}{24}\cdot {\pi }^3\right)-\frac{5}{3}\cdot {\pi }^3$

= $\frac{1}{4}+\frac{5}{24}\cdot {\pi }^3-\frac{5}{3}\cdot {\pi }^3$

= $\frac{1}{4}-\frac{35}{24}\cdot {\pi }^3$

= ${\left[-\frac{1}{6}{\left(-2x+1\right)}^{-3}\right]}_1^3$

= ${\left[-\frac{1}{6{\left(-2x+1\right)}^3}\right]}_1^3$

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

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

= $-\frac{1}{6{\left(-5\right)}^3}+\frac{1}{6{\left(-1\right)}^3}$

= $-\frac{1}{6}\cdot \left(-\frac{1}{125}\right)+\frac{1}{6}\cdot \left(-1\right)$

= $\frac{1}{750}-\frac{1}{6}$

= $\frac{1}{750}-\frac{125}{750}$

= $-\frac{62}{375}$

= ${\left[e^{-x}-\frac{4}{3}{x}^6\right]}_{-2}^{-1}$

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

= $e-\frac{4}{3}\cdot 1-\left(e^2-\frac{4}{3}\cdot 64\right)$

= $e-\frac{4}{3}-\left(e^2-\frac{256}{3}\right)$

= $-e^2-1·\left(-\frac{256}{3}\right)-\frac{4}{3}+e$

= $-e^2+\frac{256}{3}-\frac{4}{3}+e$

= $-e^2+84+e$

= ${\left[-\frac{4}{3}{\left(x-1\right)}^{\frac{3}{2}}\right]}_{17}^{26}$

= ${\left[-\frac{4}{3}{\left(\sqrt{x-1}\right)}^3\right]}_{17}^{26}$

$=$ $-\frac{4}{3}{\left(\sqrt{26-1}\right)}^3+\frac{4}{3}{\left(\sqrt{17-1}\right)}^3$

= $-\frac{4}{3}{\left(\sqrt{25}\right)}^3+\frac{4}{3}{\left(\sqrt{16}\right)}^3$

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

= $-\frac{4}{3}\cdot 125+\frac{4}{3}\cdot 64$

= $-\frac{500}{3}+\frac{256}{3}$

= $-\frac{244}{3}$