$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(2\; -\; 0)\; \cdot \; <mn>2</mn>}}{2}$ = $\frac{4}{2}$ = 2.

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

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ ( - 2 ) = 3 ⋅ ( - 2 ) = -6.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-2\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 1.5 ) = -4.5.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 2 -2 -6 -4.5 = -10.5

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

$=$ $-5^3+5^2-\left(-1^3+1^2\right)$

= $-125+25-\left(-1+1\right)$

= $-125+25-1·\left(-1\right)-1·1$

= $-125+25+1-1$

= $-100$

= ${\left[-\frac{7}{4}\cdot \sin \left(x\right)+\frac{5}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\pi }$

$=$ $-\frac{7}{4}\cdot \sin \left(\pi \right)+\frac{5}{3}\cdot \cos \left(\pi \right)-\left(-\frac{7}{4}\cdot \sin \left(\frac{1}{2}\pi \right)+\frac{5}{3}\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-\frac{7}{4}\cdot 0+\frac{5}{3}\cdot \left(-1\right)-\left(-\frac{7}{4}\cdot 1+\frac{5}{3}\cdot 0\right)$

= $0-\frac{5}{3}-\left(-\frac{7}{4}+0\right)$

= $0-\frac{5}{3}-\left(-\frac{7}{4}+0\right)$

= $-\frac{5}{3}+\frac{7}{4}$

= $-\frac{20}{12}+\frac{21}{12}$

= $\frac{1}{12}$

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

$=$ $\frac{3}{2}\cdot \sin (2\cdot \left(\frac{3}{2}\pi \right)-\pi )-\frac{3}{2}\cdot \sin (2\cdot \left(0\right)-\pi )$

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

= $\frac{3}{2}\cdot 0-\frac{3}{2}\cdot 0$

= $0+0$

= 0

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

$=$ $\frac{1}{3}{e}^{-3\cdot \pi }-\cos \left(\pi \right)-\left(\frac{1}{3}{e}^{-3\cdot \left(0\right)}-\cos \left(0\right)\right)$

= $\frac{1}{3}{e}^{-3\cdot \pi }-\left(-1\right)-\left(\frac{1}{3}{e}^0-1\right)$

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

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

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

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

= $\frac{1}{3}{e}^{-3\pi }+\frac{5}{3}$

= ${\left[\frac{1}{9}{\left(3x-7\right)}^3-\frac{5}{2}{x}^2\right]}_1^2$

$=$ $\frac{1}{9}\cdot {\left(3\cdot 2-7\right)}^3-\frac{5}{2}\cdot 2^2-\left(\frac{1}{9}\cdot {\left(3\cdot 1-7\right)}^3-\frac{5}{2}\cdot 1^2\right)$

= $\frac{1}{9}\cdot {\left(6-7\right)}^3-\frac{5}{2}\cdot 4-\left(\frac{1}{9}\cdot {\left(3-7\right)}^3-\frac{5}{2}\cdot 1\right)$

= $\frac{1}{9}\cdot {\left(-1\right)}^3-10-\left(\frac{1}{9}\cdot {\left(-4\right)}^3-\frac{5}{2}\right)$

= $\frac{1}{9}\cdot \left(-1\right)-10-\left(\frac{1}{9}\cdot \left(-64\right)-\frac{5}{2}\right)$

= $-\frac{1}{9}-10-\left(-\frac{64}{9}-\frac{5}{2}\right)$

= $-\frac{1}{9}-\frac{90}{9}-\left(-\frac{128}{18}-\frac{45}{18}\right)$

= $-\frac{91}{9}-1·\left(-\frac{173}{18}\right)$

= $-\frac{91}{9}+\frac{173}{18}$

= $-\frac{182}{18}+\frac{173}{18}$

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