$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ 4 = 3 ⋅ 4 = 12.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = 12 +4 -3 = 13

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

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

= $-\frac{3}{2}\cdot 9+6-\left(-\frac{3}{2}\cdot 1-2\right)$

= $-\frac{27}{2}+6-\left(-\frac{3}{2}-2\right)$

= $-\frac{27}{2}+\frac{12}{2}-\left(-\frac{3}{2}-\frac{4}{2}\right)$

= $-\frac{15}{2}-1·\left(-\frac{7}{2}\right)$

= $-\frac{15}{2}+\frac{7}{2}$

= $-4$

= ${\left[-3\cdot \sin \left(x\right)-\frac{8}{9}{x}^{\frac{3}{2}}\right]}_0^1$

= ${\left[-3\cdot \sin \left(x\right)-\frac{8}{9}{\left(\sqrt{x}\right)}^3\right]}_0^1$

$=$ $-3\cdot \sin \left(1\right)-\frac{8}{9}{\left(\sqrt{1}\right)}^3-\left(-3\cdot \sin \left(0\right)-\frac{8}{9}{\left(\sqrt{0}\right)}^3\right)$

= $-3\cdot \sin \left(1\right)-\frac{8}{9}\cdot 1^3-\left(-3\cdot 0-\frac{8}{9}\cdot 0^3\right)$

= $-3\cdot \sin \left(1\right)-\frac{8}{9}\cdot 1-\left(0-\frac{8}{9}\cdot 0\right)$

= $-3\cdot \sin \left(1\right)-\frac{8}{9}-\left(0+0\right)$

= $-3\cdot \sin \left(1\right)-\frac{8}{9}+0$

= $-3\cdot \sin \left(1\right)-\frac{8}{9}$

= ${\left[\frac{2}{3}{\left(-2x+1\right)}^{\frac{3}{2}}\right]}_{-\frac{15}{2}}^0$

= ${\left[\frac{2}{3}{\left(\sqrt{-2x+1}\right)}^3\right]}_{-\frac{15}{2}}^0$

$=$ $\frac{2}{3}{\left(\sqrt{-2\cdot \left(0\right)+1}\right)}^3-\frac{2}{3}{\left(\sqrt{-2\cdot \left(-\frac{15}{2}\right)+1}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{0+1}\right)}^3-\frac{2}{3}{\left(\sqrt{15+1}\right)}^3$

= $\frac{2}{3}{\left(\sqrt{1}\right)}^3-\frac{2}{3}{\left(\sqrt{16}\right)}^3$

= $\frac{2}{3}\cdot 1^3-\frac{2}{3}\cdot 4^3$

= $\frac{2}{3}\cdot 1-\frac{2}{3}\cdot 64$

= $\frac{2}{3}-\frac{128}{3}$

= $-42$

= ${\left[-\frac{9}{2}{e}^{-x}-\cos \left(x\right)\right]}_0^{\pi }$

$=$ $-\frac{9}{2}{e}^{-\pi }-\cos \left(\pi \right)-\left(-\frac{9}{2}{e}^{-\left(0\right)}-\cos \left(0\right)\right)$

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

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

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

= $-\frac{9}{2}{e}^{-\pi }+1-1·\left(-\frac{11}{2}\right)$

= $-\frac{9}{2}{e}^{-\pi }+1+\frac{11}{2}$

= $-\frac{9}{2}{e}^{-\pi }+\frac{13}{2}$

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

$=$ $\frac{1}{2}{e}^{2\cdot 3-2}-\frac{1}{2}{e}^{2\cdot 2-2}$

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

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