$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-9}}{2}$ = -4.5.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ ( - 3 ) = 2 ⋅ ( - 3 ) = -6.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-3\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 3.5 ) = -10.5.

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 -4.5 -6 -10.5 = -18

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

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

= $-\frac{1}{3}\cdot 27-15-\left(-\frac{1}{3}\cdot 1-5\right)$

= $-9-15-\left(-\frac{1}{3}-5\right)$

= $-24-\left(-\frac{1}{3}-\frac{15}{3}\right)$

= $-24-1·\left(-\frac{16}{3}\right)$

= $-24+\frac{16}{3}$

= $-\frac{72}{3}+\frac{16}{3}$

= $-\frac{56}{3}$

= ${\left[-\frac{7}{12}{x}^4-\cos \left(x\right)\right]}_0^{\pi }$

$=$ $-\frac{7}{12}\cdot {\pi }^4-\cos \left(\pi \right)-\left(-\frac{7}{12}\cdot {\left(0\right)}^4-\cos \left(0\right)\right)$

= $-\frac{7}{12}\cdot {\pi }^4-\left(-1\right)-\left(-\frac{7}{12}\cdot 0-1\right)$

= $-\frac{7}{12}\cdot {\pi }^4+1-\left(0-1\right)$

= $1-\frac{7}{12}\cdot {\pi }^4+1$

= $1+1-\frac{7}{12}\cdot {\pi }^4$

= $2-\frac{7}{12}\cdot {\pi }^4$

= ${\left[\frac{1}{3}{e}^{-3x+4}\right]}_1^2$

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

= $\frac{1}{3}{e}^{-6+4}-\frac{1}{3}{e}^{-3+4}$

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

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

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

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

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

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

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

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

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

= $2+\frac{3}{2\pi }+2-\frac{3}{\pi }$

= $2+2+\frac{3}{2\pi }-\frac{3}{\pi }$

= $4-\frac{3}{2\pi }$

= ${\left[-\frac{1}{8}{\left(-2x+4\right)}^4+\frac{3}{2}{x}^2\right]}_1^4$

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

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

= $-\frac{1}{8}\cdot {\left(-4\right)}^4+24-\left(-\frac{1}{8}\cdot 2^4+\frac{3}{2}\right)$

= $-\frac{1}{8}\cdot 256+24-\left(-\frac{1}{8}\cdot 16+\frac{3}{2}\right)$

= $-32+24-\left(-2+\frac{3}{2}\right)$

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

= $-8-1·\left(-\frac{1}{2}\right)$

= $-8+\frac{1}{2}$

= $-\frac{16}{2}+\frac{1}{2}$

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