$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₂ = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₄ = $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (9 - 7) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 3 ) = -6.

Somit gilt:

$$\underset{0}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{9}{\int }}\mathrm{f(x)}dx$$ = 4 -6 -8 -6 = -16

= ${\left[\frac{5}{3}{x}^3-5x\right]}_{-2}^{-1}$

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

= $\frac{5}{3}\cdot \left(-1\right)+5-\left(\frac{5}{3}\cdot \left(-8\right)+10\right)$

= $-\frac{5}{3}+5-\left(-\frac{40}{3}+10\right)$

= $-\frac{5}{3}+\frac{15}{3}-\left(-\frac{40}{3}+\frac{30}{3}\right)$

= $\frac{10}{3}-1·\left(-\frac{10}{3}\right)$

= $\frac{10}{3}+\frac{10}{3}$

= $\frac{20}{3}$

= ${\left[-\frac{1}{3}{x}^{-1}-\frac{7}{2}{e}^x\right]}_2^4$

= ${\left[-\frac{1}{3x}-\frac{7}{2}{e}^x\right]}_2^4$

$=$ $-\frac{1}{3\cdot 4}-\frac{7}{2}{e}^4-\left(-\frac{1}{3\cdot 2}-\frac{7}{2}{e}^2\right)$

= $-\frac{1}{3}\cdot \left(\frac{1}{4}\right)-\frac{7}{2}{e}^4-\left(-\frac{1}{3}\cdot \left(\frac{1}{2}\right)-\frac{7}{2}{e}^2\right)$

= $-\frac{1}{12}-\frac{7}{2}{e}^4-\left(-\frac{1}{6}-\frac{7}{2}{e}^2\right)$

= $-\frac{7}{2}{e}^4-\frac{1}{12}-\left(-\frac{7}{2}{e}^2-\frac{1}{6}\right)$

= $-\frac{7}{2}{e}^4-\frac{1}{12}+\frac{7}{2}{e}^2-1·\left(-\frac{1}{6}\right)$

= $-\frac{7}{2}{e}^4-\frac{1}{12}+\frac{7}{2}{e}^2+\frac{1}{6}$

= $-\frac{7}{2}{e}^4+\frac{7}{2}{e}^2-\frac{1}{12}+\frac{1}{6}$

= $-\frac{7}{2}{e}^4+\frac{7}{2}{e}^2+\frac{1}{12}$

= ${\left[\frac{2}{3}{\left(-x+1\right)}^3+\frac{5}{2}{x}^2\right]}_1^2$

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

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

= $\frac{2}{3}\cdot \left(-1\right)+10-\left(\frac{2}{3}\cdot 0+\frac{5}{2}\right)$

= $-\frac{2}{3}+10-\left(0+\frac{5}{2}\right)$

= $-\frac{2}{3}+\frac{30}{3}-\left(0+\frac{5}{2}\right)$

= $\frac{28}{3}-\frac{5}{2}$

= $\frac{56}{6}-\frac{15}{6}$

= $\frac{41}{6}$

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

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

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

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

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

= $5+\frac{5}{2}$

= $\frac{10}{2}+\frac{5}{2}$

= $\frac{15}{2}$

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

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

= $-e^{10-3}+e^{4-3}$

= $-e^7+e$

= $-e^7+e$