$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(4\; -\; 2)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (6 - 4) ⋅ 4 = 2 ⋅ 4 = 8.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (8 - 6) ⋅ $\frac{\mathrm{4\; +\; <mn>3</mn>}}{2}$ = 2 ⋅ 3.5 = 7.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₅ = (10 - 8) ⋅ 3 = 2 ⋅ 3 = 6.

Somit gilt:

$$\underset{2}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ + I₅ = $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4 +8 +7 +6 = 25

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

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

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

= $\frac{27}{2}-9-\left(\frac{3}{2}-3\right)$

= $\frac{27}{2}-\frac{18}{2}-\left(\frac{3}{2}-\frac{6}{2}\right)$

= $\frac{9}{2}-1·\left(-\frac{3}{2}\right)$

= $\frac{9}{2}+\frac{3}{2}$

= $6$

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

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

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

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

= $\frac{3}{4}+\frac{9}{4}\cdot {\pi }^3-\left(-\frac{3}{4}+\frac{1}{12}\cdot {\pi }^3\right)$

= $\frac{3}{4}+\frac{9}{4}\cdot {\pi }^3-1·\left(-\frac{3}{4}\right)-1·\frac{1}{12}\cdot {\pi }^3$

= $\frac{3}{4}+\frac{9}{4}\cdot {\pi }^3+\frac{3}{4}-\frac{1}{12}\cdot {\pi }^3$

= $\frac{3}{4}+\frac{3}{4}+\frac{9}{4}\cdot {\pi }^3-\frac{1}{12}\cdot {\pi }^3$

= $\frac{3}{2}+\frac{13}{6}\cdot {\pi }^3$

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

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

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

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

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

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

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

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

$=$ $5\cdot \cos \left(\frac{3}{2}\pi \right)+\frac{1}{2}{e}^{2\cdot \left(\frac{3}{2}\pi \right)}-\left(5\cdot \cos \left(\frac{1}{2}\pi \right)+\frac{1}{2}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $5\cdot 0+\frac{1}{2}{e}^{2\cdot \left(\frac{3}{2}\pi \right)}-\left(5\cdot 0+\frac{1}{2}{e}^{2\cdot \left(\frac{1}{2}\pi \right)}\right)$

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

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

= ${\left[-\frac{1}{3}{\left(2x-1\right)}^3-3x\right]}_2^4$

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

= $-\frac{1}{3}\cdot {\left(8-1\right)}^3-12-\left(-\frac{1}{3}\cdot {\left(4-1\right)}^3-6\right)$

= $-\frac{1}{3}\cdot 7^3-12-\left(-\frac{1}{3}\cdot 3^3-6\right)$

= $-\frac{1}{3}\cdot 343-12-\left(-\frac{1}{3}\cdot 27-6\right)$

= $-\frac{343}{3}-12-\left(-9-6\right)$

= $-\frac{343}{3}-\frac{36}{3}-1·\left(-15\right)$

= $-\frac{379}{3}+15$

= $-\frac{379}{3}+\frac{45}{3}$

= $-\frac{379}{3}+15$

= $-\frac{379}{3}+\frac{45}{3}$

= $-\frac{334}{3}$