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

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 = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 2 ) = 2 ⋅ ( - 2 ) = -4.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-2\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 2.5 ) = -7.5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 3 -3 -4 -7.5 = -11.5

= ${\left[\frac{1}{2}{x}^2+4x\right]}_{-3}^0$

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

= $\frac{1}{2}\cdot 0+0-\left(\frac{1}{2}\cdot 9-12\right)$

= $0+0-\left(\frac{9}{2}-12\right)$

= $0-\left(\frac{9}{2}-\frac{24}{2}\right)$

= $-1·\left(-\frac{15}{2}\right)$

= $\frac{15}{2}$

= ${\left[\frac{3}{2}{e}^{-2x}+\frac{5}{3}{x}^{-3}\right]}_1^2$

= ${\left[\frac{3}{2}{e}^{-2x}+\frac{5}{3x^3}\right]}_1^2$

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

= $\frac{3}{2}{e}^{-4}+\frac{5}{3}\cdot \left(\frac{1}{8}\right)-\left(\frac{3}{2}{e}^{-2}+\frac{5}{3}\cdot 1\right)$

= $\frac{3}{2}{e}^{-4}+\frac{5}{24}-\left(\frac{3}{2}{e}^{-2}+\frac{5}{3}\right)$

= $-\frac{3}{2}{e}^{-2}-1·\frac{5}{3}+\frac{3}{2}{e}^{-4}+\frac{5}{24}$

= $-\frac{3}{2}{e}^{-2}-\frac{5}{3}+\frac{3}{2}{e}^{-4}+\frac{5}{24}$

= $-\frac{3}{2}{e}^{-2}+\frac{3}{2}{e}^{-4}-\frac{5}{3}+\frac{5}{24}$

= $-\frac{3}{2}{e}^{-2}+\frac{3}{2}{e}^{-4}-\frac{35}{24}$

= ${\left[\frac{3}{8}{\left(-2x+5\right)}^4+\frac{3}{2}{x}^2\right]}_0^2$

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

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

= $\frac{3}{8}\cdot 1^4+6-\left(\frac{3}{8}\cdot 5^4+0\right)$

= $\frac{3}{8}\cdot 1+6-\left(\frac{3}{8}\cdot 625+0\right)$

= $\frac{3}{8}+6-\left(\frac{1875}{8}+0\right)$

= $\frac{3}{8}+\frac{48}{8}-\left(\frac{1875}{8}+0\right)$

= $\frac{51}{8}-\frac{1875}{8}$

= $-228$

= ${\left[-\frac{8}{9}{x}^{\frac{3}{2}}+\frac{1}{2}{e}^{-x}\right]}_4^{16}$

= ${\left[-\frac{8}{9}{\left(\sqrt{x}\right)}^3+\frac{1}{2}{e}^{-x}\right]}_4^{16}$

$=$ $-\frac{8}{9}{\left(\sqrt{16}\right)}^3+\frac{1}{2}{e}^{-16}-\left(-\frac{8}{9}{\left(\sqrt{4}\right)}^3+\frac{1}{2}{e}^{-4}\right)$

= $-\frac{8}{9}\cdot 4^3+\frac{1}{2}{e}^{-16}-\left(-\frac{8}{9}\cdot 2^3+\frac{1}{2}{e}^{-4}\right)$

= $-\frac{8}{9}\cdot 64+\frac{1}{2}{e}^{-16}-\left(-\frac{8}{9}\cdot 8+\frac{1}{2}{e}^{-4}\right)$

= $-\frac{512}{9}+\frac{1}{2}{e}^{-16}-\left(-\frac{64}{9}+\frac{1}{2}{e}^{-4}\right)$

= $\frac{1}{2}{e}^{-16}-\frac{512}{9}-\left(\frac{1}{2}{e}^{-4}-\frac{64}{9}\right)$

= $-\frac{1}{2}{e}^{-4}-1·\left(-\frac{64}{9}\right)+\frac{1}{2}{e}^{-16}-\frac{512}{9}$

= $-\frac{1}{2}{e}^{-4}+\frac{64}{9}+\frac{1}{2}{e}^{-16}-\frac{512}{9}$

= $-\frac{1}{2}{e}^{-4}+\frac{1}{2}{e}^{-16}+\frac{64}{9}-\frac{512}{9}$

= $-\frac{1}{2}{e}^{-4}+\frac{1}{2}{e}^{-16}-\frac{448}{9}$

= ${\left[\frac{1}{4}{\left(3x-7\right)}^4\right]}_1^3$

$=$ $\frac{1}{4}\cdot {\left(3\cdot 3-7\right)}^4-\frac{1}{4}\cdot {\left(3\cdot 1-7\right)}^4$

= $\frac{1}{4}\cdot {\left(9-7\right)}^4-\frac{1}{4}\cdot {\left(3-7\right)}^4$

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

= $\frac{1}{4}\cdot 16-\frac{1}{4}\cdot 256$

= $4-64$

= $-60$