$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ 4 = 3 ⋅ 4 = 12.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{4\; +\; <mn>5</mn>}}{2}$ = 3 ⋅ 4.5 = 13.5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 +4 +12 +13.5 = 26.5

= ${\left[\frac{5}{3}{x}^3+2x^2\right]}_0^2$

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

= $\frac{5}{3}\cdot 8+2\cdot 4-\left(\frac{5}{3}\cdot 0+2\cdot 0\right)$

= $\frac{40}{3}+8-\left(0+0\right)$

= $\frac{40}{3}+\frac{24}{3}+0$

= $\frac{64}{3}$

= ${\left[\frac{4}{3}{x}^{\frac{3}{2}}+\frac{1}{3}{e}^{-3x}\right]}_4^{16}$

= ${\left[\frac{4}{3}{\left(\sqrt{x}\right)}^3+\frac{1}{3}{e}^{-3x}\right]}_4^{16}$

$=$ $\frac{4}{3}{\left(\sqrt{16}\right)}^3+\frac{1}{3}{e}^{-3\cdot 16}-\left(\frac{4}{3}{\left(\sqrt{4}\right)}^3+\frac{1}{3}{e}^{-3\cdot 4}\right)$

= $\frac{4}{3}\cdot 4^3+\frac{1}{3}{e}^{-48}-\left(\frac{4}{3}\cdot 2^3+\frac{1}{3}{e}^{-12}\right)$

= $\frac{4}{3}\cdot 64+\frac{1}{3}{e}^{-48}-\left(\frac{4}{3}\cdot 8+\frac{1}{3}{e}^{-12}\right)$

= $\frac{256}{3}+\frac{1}{3}{e}^{-48}-\left(\frac{32}{3}+\frac{1}{3}{e}^{-12}\right)$

= $\frac{1}{3}{e}^{-48}+\frac{256}{3}-\left(\frac{1}{3}{e}^{-12}+\frac{32}{3}\right)$

= $-\frac{1}{3}{e}^{-12}-1·\frac{32}{3}+\frac{1}{3}{e}^{-48}+\frac{256}{3}$

= $-\frac{1}{3}{e}^{-12}-\frac{32}{3}+\frac{1}{3}{e}^{-48}+\frac{256}{3}$

= $-\frac{1}{3}{e}^{-12}+\frac{1}{3}{e}^{-48}-\frac{32}{3}+\frac{256}{3}$

= $-\frac{1}{3}{e}^{-12}+\frac{1}{3}{e}^{-48}+\frac{224}{3}$

= ${\left[-\frac{1}{3}{\left(-x+3\right)}^{-3}\right]}_5^7$

= ${\left[-\frac{1}{3{\left(-x+3\right)}^3}\right]}_5^7$

$=$ $-\frac{1}{3{\left(-7+3\right)}^3}+\frac{1}{3{\left(-5+3\right)}^3}$

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

= $-\frac{1}{3}\cdot \left(-\frac{1}{64}\right)+\frac{1}{3}\cdot \left(-\frac{1}{8}\right)$

= $\frac{1}{192}-\frac{1}{24}$

= $\frac{1}{192}-\frac{8}{192}$

= $-\frac{7}{192}$

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

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

= $e^{-2}-\frac{1}{3}\cdot 64-\left(e^{-3}-\frac{1}{3}\cdot 729\right)$

= $e^{-2}-\frac{64}{3}-\left(e^{-3}-243\right)$

= $e^{-2}-\frac{64}{3}-e^{-3}-1·\left(-243\right)$

= $e^{-2}-\frac{64}{3}-e^{-3}+243$

= $e^{-2}-e^{-3}-\frac{64}{3}+243$

= $e^{-2}-e^{-3}+\frac{665}{3}$

= ${\left[\frac{1}{3}{e}^{3x-4}\right]}_0^1$

$=$ $\frac{1}{3}{e}^{3\cdot 1-4}-\frac{1}{3}{e}^{3\cdot 0-4}$

= $\frac{1}{3}{e}^{3-4}-\frac{1}{3}{e}^{0-4}$

= $\frac{1}{3}{e}^{-1}-\frac{1}{3}{e}^{-4}$