$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>4</mn>}}{2}$ = $\frac{8}{2}$ = 4.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ 4 = 3 ⋅ 4 = 12.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{4\; +\; <mn>5</mn>}}{2}$ = 2 ⋅ 4.5 = 9.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 4 +12 +9 = 25

= ${\left[x^3-x^2\right]}_{-1}^1$

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

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

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

= $1-1+1+1$

= $2$

= ${\left[-\frac{2}{3}{x}^{\frac{3}{2}}-x^4\right]}_4^9$

= ${\left[-\frac{2}{3}{\left(\sqrt{x}\right)}^3-x^4\right]}_4^9$

$=$ $-\frac{2}{3}{\left(\sqrt{9}\right)}^3-9^4-\left(-\frac{2}{3}{\left(\sqrt{4}\right)}^3-4^4\right)$

= $-\frac{2}{3}\cdot 3^3-6561-\left(-\frac{2}{3}\cdot 2^3-256\right)$

= $-\frac{2}{3}\cdot 27-6561-\left(-\frac{2}{3}\cdot 8-256\right)$

= $-18-6561-\left(-\frac{16}{3}-256\right)$

= $-6579-\left(-\frac{16}{3}-\frac{768}{3}\right)$

= $-6579-1·\left(-\frac{784}{3}\right)$

= $-6579+\frac{784}{3}$

= $-\frac{19737}{3}+\frac{784}{3}$

= $-6579+\frac{784}{3}$

= $-\frac{19737}{3}+\frac{784}{3}$

= $-\frac{18953}{3}$

= ${\left[\frac{3}{4}{\left(x-1\right)}^4\right]}_0^2$

$=$ $\frac{3}{4}\cdot {\left(2-1\right)}^4-\frac{3}{4}\cdot {\left(0-1\right)}^4$

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

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

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

= 0

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

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

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

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

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

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

= $\frac{1}{3}{e}^{\frac{3}{2}\pi }+\frac{7}{6}$

= ${\left[\frac{1}{3}{\left(3x-7\right)}^3+x^2\right]}_1^4$

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

= $\frac{1}{3}\cdot {\left(12-7\right)}^3+16-\left(\frac{1}{3}\cdot {\left(3-7\right)}^3+1\right)$

= $\frac{1}{3}\cdot 5^3+16-\left(\frac{1}{3}\cdot {\left(-4\right)}^3+1\right)$

= $\frac{1}{3}\cdot 125+16-\left(\frac{1}{3}\cdot \left(-64\right)+1\right)$

= $\frac{125}{3}+16-\left(-\frac{64}{3}+1\right)$

= $\frac{125}{3}+\frac{48}{3}-\left(-\frac{64}{3}+\frac{3}{3}\right)$

= $\frac{173}{3}-1·\left(-\frac{61}{3}\right)$

= $\frac{173}{3}+\frac{61}{3}$

= $78$