$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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:

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

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ ( - 4 ) = 3 ⋅ ( - 4 ) = -12.

I₄ = $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 7) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>1</mn>\; <mo>)</mo>}}{2}$ = 3 ⋅ ( - 2.5 ) = -7.5.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -4 -12 -7.5 = -23.5

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

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

= $-\frac{4}{3}\cdot 64-\frac{3}{2}\cdot 16-\left(-\frac{4}{3}\cdot 1-\frac{3}{2}\cdot 1\right)$

= $-\frac{256}{3}-24-\left(-\frac{4}{3}-\frac{3}{2}\right)$

= $-\frac{256}{3}-\frac{72}{3}-\left(-\frac{8}{6}-\frac{9}{6}\right)$

= $-\frac{328}{3}-1·\left(-\frac{17}{6}\right)$

= $-\frac{328}{3}+\frac{17}{6}$

= $-\frac{656}{6}+\frac{17}{6}$

= $-\frac{213}{2}$

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

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

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

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

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

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

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

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

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

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

= $e^2-e$

= $e^2-e$

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

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

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

= $-\frac{7}{2}\cdot 0-\frac{1}{3{\left(2\pi \right)}^3}-\left(-\frac{7}{2}\cdot 0-\frac{1}{3{\pi }^3}\right)$

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

= $-\frac{1}{24{\pi }^3}+\frac{1}{3{\pi }^3}$

= $-\frac{1}{24{\pi }^3}+\frac{8}{24{\pi }^3}$

= $\frac{7}{24{\pi }^3}$

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

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

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

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

= $-\frac{1}{4}\cdot 16+12-\left(-\frac{1}{4}\cdot 1+3\right)$

= $-4+12-\left(-\frac{1}{4}+3\right)$

= $8-\left(-\frac{1}{4}+\frac{12}{4}\right)$

= $8-1·\frac{11}{4}$

= $8-\frac{11}{4}$

= $\frac{32}{4}-\frac{11}{4}$

= $\frac{21}{4}$