$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (3 - 0) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

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

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (10 - 7) ⋅ 2 = 3 ⋅ 2 = 6.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -6 -2 +2 +6 = 0

= ${\left[-\frac{1}{2}{x}^2-4x\right]}_1^2$

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

= $-\frac{1}{2}\cdot 4-8-\left(-\frac{1}{2}\cdot 1-4\right)$

= $-2-8-\left(-\frac{1}{2}-4\right)$

= $-10-\left(-\frac{1}{2}-\frac{8}{2}\right)$

= $-10-1·\left(-\frac{9}{2}\right)$

= $-10+\frac{9}{2}$

= $-\frac{20}{2}+\frac{9}{2}$

= $-\frac{11}{2}$

= ${\left[2x^{\frac{3}{2}}+\frac{7}{9}{e}^{-3x}\right]}_1^{16}$

= ${\left[2{\left(\sqrt{x}\right)}^3+\frac{7}{9}{e}^{-3x}\right]}_1^{16}$

$=$ $2{\left(\sqrt{16}\right)}^3+\frac{7}{9}{e}^{-3\cdot 16}-\left(2{\left(\sqrt{1}\right)}^3+\frac{7}{9}{e}^{-3\cdot 1}\right)$

= $2\cdot 4^3+\frac{7}{9}{e}^{-48}-\left(2\cdot 1^3+\frac{7}{9}{e}^{-3}\right)$

= $2\cdot 64+\frac{7}{9}{e}^{-48}-\left(2\cdot 1+\frac{7}{9}{e}^{-3}\right)$

= $128+\frac{7}{9}{e}^{-48}-\left(2+\frac{7}{9}{e}^{-3}\right)$

= $\frac{7}{9}{e}^{-48}+128-\left(\frac{7}{9}{e}^{-3}+2\right)$

= $-\frac{7}{9}{e}^{-3}-1·2+\frac{7}{9}{e}^{-48}+128$

= $-\frac{7}{9}{e}^{-3}-2+\frac{7}{9}{e}^{-48}+128$

= $-\frac{7}{9}{e}^{-3}+\frac{7}{9}{e}^{-48}-2+128$

= $-\frac{7}{9}{e}^{-3}+\frac{7}{9}{e}^{-48}+126$

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

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

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

= $-\frac{1}{2}\cdot 0+\frac{1}{2}\cdot 0$

= $0+0$

= 0

= ${\left[-4\cdot \sin \left(x\right)+\frac{1}{12}{x}^6\right]}_0^{\frac{3}{2}\pi }$

$=$ $-4\cdot \sin \left(\frac{3}{2}\pi \right)+\frac{1}{12}\cdot {\left(\frac{3}{2}\pi \right)}^6-\left(-4\cdot \sin \left(0\right)+\frac{1}{12}\cdot {\left(0\right)}^6\right)$

= $-4\cdot \left(-1\right)+\frac{1}{12}\cdot {\left(\frac{3}{2}\pi \right)}^6-\left(-4\cdot 0+\frac{1}{12}\cdot 0\right)$

= $4+\frac{1}{12}\cdot {\left(\frac{3}{2}\pi \right)}^6-\left(0+0\right)$

= $4+\frac{243}{256}\cdot {\pi }^6+0$

= $4+\frac{243}{256}\cdot {\pi }^6$

= ${\left[-\cos \left(x+\frac{1}{2}\pi \right)\right]}_0^{\frac{3}{2}\pi }$

$=$ $-\cos \left(\frac{3}{2}\pi +\frac{1}{2}\pi \right)+\cos \left(0+\frac{1}{2}\pi \right)$

= $-\cos \left(2\pi \right)+\cos \left(\frac{1}{2}\pi \right)$

= $-1+0$

= $-1$