$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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:

I₂ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 4) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

Somit gilt:

$$\underset{2}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -3 -9 = -12

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

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

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

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

= $\frac{16}{3}+\frac{24}{3}-\left(\frac{2}{3}+\frac{12}{3}\right)$

= $\frac{40}{3}-1·\frac{14}{3}$

= $\frac{40}{3}-\frac{14}{3}$

= $\frac{26}{3}$

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

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

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

= $6+e^{\frac{1}{2}\pi }-\left(0+1\right)$

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

= $e^{\frac{1}{2}\pi }+5$

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

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

= $\cos \left(-2\pi \right)-\cos (-\pi )$

= $1-\left(-1\right)$

= $1+1$

= $2$

= ${\left[-\frac{8}{3}{x}^{\frac{3}{2}}+\frac{7}{2}\cdot \sin \left(x\right)\right]}_1^4$

= ${\left[-\frac{8}{3}{\left(\sqrt{x}\right)}^3+\frac{7}{2}\cdot \sin \left(x\right)\right]}_1^4$

$=$ $-\frac{8}{3}{\left(\sqrt{4}\right)}^3+\frac{7}{2}\cdot \sin \left(4\right)-\left(-\frac{8}{3}{\left(\sqrt{1}\right)}^3+\frac{7}{2}\cdot \sin \left(1\right)\right)$

= $-\frac{8}{3}\cdot 2^3+\frac{7}{2}\cdot \sin \left(4\right)-\left(-\frac{8}{3}\cdot 1^3+\frac{7}{2}\cdot \sin \left(1\right)\right)$

= $-\frac{8}{3}\cdot 8+\frac{7}{2}\cdot \sin \left(4\right)-\left(-\frac{8}{3}\cdot 1+\frac{7}{2}\cdot \sin \left(1\right)\right)$

= $-\frac{64}{3}+\frac{7}{2}\cdot \sin \left(4\right)-\left(-\frac{8}{3}+\frac{7}{2}\cdot \sin \left(1\right)\right)$

= $\frac{7}{2}\cdot \sin \left(4\right)-\frac{64}{3}-\left(\frac{7}{2}\cdot \sin \left(1\right)-\frac{8}{3}\right)$

= $\frac{7}{2}\cdot \sin \left(4\right)-\frac{64}{3}-1·\frac{7}{2}\cdot \sin \left(1\right)-1·\left(-\frac{8}{3}\right)$

= $\frac{7}{2}\cdot \sin \left(4\right)-\frac{64}{3}-\frac{7}{2}\cdot \sin \left(1\right)+\frac{8}{3}$

= $\frac{7}{2}\cdot \sin \left(4\right)-\frac{7}{2}\cdot \sin \left(1\right)-\frac{64}{3}+\frac{8}{3}$

= $\frac{7}{2}\cdot \sin \left(4\right)-\frac{7}{2}\cdot \sin \left(1\right)-\frac{56}{3}$

= ${\left[\frac{3}{4}{\left(x-1\right)}^4\right]}_0^1$

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

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

= $\frac{3}{4}\cdot 0-\frac{3}{4}\cdot 1$

= $0-\frac{3}{4}$

= $0-\frac{3}{4}$

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