$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₁ = $\frac{\mathrm{(3\; -\; 0)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₂ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{2}{2}$ = 1.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (7 - 5) ⋅ 1 = 2 ⋅ 1 = 2.

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

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ = $$\underset{0}{\overset{3}{\int }}\mathrm{f(x)}dx$$ + $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ + $$\underset{7}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -6 +1 +2 +4.5 = 1.5

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

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

= $\frac{4}{3}\cdot \left(-1\right)+\frac{1}{2}\cdot 1-\left(\frac{4}{3}\cdot \left(-27\right)+\frac{1}{2}\cdot 9\right)$

= $-\frac{4}{3}+\frac{1}{2}-\left(-36+\frac{9}{2}\right)$

= $-\frac{8}{6}+\frac{3}{6}-\left(-\frac{72}{2}+\frac{9}{2}\right)$

= $-\frac{5}{6}-1·\left(-\frac{63}{2}\right)$

= $-\frac{5}{6}+\frac{63}{2}$

= $-\frac{5}{6}+\frac{189}{6}$

= $\frac{92}{3}$

= ${\left[-\frac{1}{20}{x}^5-3x^{\frac{4}{3}}\right]}_1^4$

= ${\left[-\frac{1}{20}{x}^5-3{\left(\sqrt[3]{x}\right)}^4\right]}_1^4$

$=$ $-\frac{1}{20}\cdot 4^5-3{\left(\sqrt[3]{4}\right)}^4-\left(-\frac{1}{20}\cdot 1^5-3{\left(\sqrt[3]{1}\right)}^4\right)$

= $-\frac{1}{20}\cdot 1024-3\cdot 1^4-\left(-\frac{1}{20}\cdot 1-3\cdot 1^4\right)$

= $-\frac{256}{5}-3\cdot 1-\left(-\frac{1}{20}-3\cdot 1\right)$

= $-\frac{256}{5}-3-\left(-\frac{1}{20}-3\right)$

= $-\frac{256}{5}-\frac{15}{5}-\left(-\frac{1}{20}-\frac{60}{20}\right)$

= $-\frac{271}{5}-1·\left(-\frac{61}{20}\right)$

= $-\frac{271}{5}+\frac{61}{20}$

= $-\frac{1084}{20}+\frac{61}{20}$

= $-\frac{271}{5}+\frac{61}{20}$

= $-\frac{1084}{20}+\frac{61}{20}$

= $-\frac{1023}{20}$

= ${\left[\frac{1}{2}{\left(x-3\right)}^4-3x\right]}_2^5$

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

= $\frac{1}{2}\cdot 2^4-15-\left(\frac{1}{2}\cdot {\left(-1\right)}^4-6\right)$

= $\frac{1}{2}\cdot 16-15-\left(\frac{1}{2}\cdot 1-6\right)$

= $8-15-\left(\frac{1}{2}-6\right)$

= $-7-\left(\frac{1}{2}-\frac{12}{2}\right)$

= $-7-1·\left(-\frac{11}{2}\right)$

= $-7+\frac{11}{2}$

= $-\frac{14}{2}+\frac{11}{2}$

= $-7+\frac{11}{2}$

= $-\frac{14}{2}+\frac{11}{2}$

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

= ${\left[-4x^{\frac{1}{2}}+e^{-3x}\right]}_1^9$

= ${\left[-4\sqrt{x}+e^{-3x}\right]}_1^9$

$=$ $-4\sqrt{9}+e^{-3\cdot 9}-\left(-4\sqrt{1}+e^{-3\cdot 1}\right)$

= $-4\cdot 3+e^{-27}-\left(-4\cdot 1+e^{-3}\right)$

= $-12+e^{-27}-\left(-4+e^{-3}\right)$

= $e^{-27}-12-\left(e^{-3}-4\right)$

= $-e^{-3}-1·\left(-4\right)+e^{-27}-12$

= $-e^{-3}+4+e^{-27}-12$

= $-e^{-3}+e^{-27}+4-12$

= $-e^{-3}+e^{-27}-8$

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

$=$ $3\cdot \sin (\frac{1}{2}\pi -\pi )-3\cdot \sin (0-\pi )$

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

= $3\cdot \left(-1\right)-3\cdot 0$

= $-3+0$

= $-3$