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

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(8\; -\; 5)\; \cdot \; <mn>1</mn>}}{2}$ = $\frac{3}{2}$ = 1.5.

Somit gilt:

$$\underset{3}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ = -4 +1.5 = -2.5

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

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

= $-\left(-1\right)-1-\left(-\left(-27\right)-9\right)$

= $1-1-\left(27-9\right)$

= $0-1·18$

= $0-18$

= $-18$

= ${\left[\frac{16}{15}{x}^{\frac{5}{4}}-\frac{7}{12}{e}^{3x}\right]}_1^4$

= ${\left[\frac{16}{15}{\left(\sqrt[4]{x}\right)}^5-\frac{7}{12}{e}^{3x}\right]}_1^4$

$=$ $\frac{16}{15}{\left(\sqrt[4]{4}\right)}^5-\frac{7}{12}{e}^{3\cdot 4}-\left(\frac{16}{15}{\left(\sqrt[4]{1}\right)}^5-\frac{7}{12}{e}^{3\cdot 1}\right)$

= $\frac{16}{15}\cdot 1^5-\frac{7}{12}{e}^{12}-\left(\frac{16}{15}\cdot 1^5-\frac{7}{12}{e}^3\right)$

= $\frac{16}{15}\cdot 1-\frac{7}{12}{e}^{12}-\left(\frac{16}{15}\cdot 1-\frac{7}{12}{e}^3\right)$

= $\frac{16}{15}-\frac{7}{12}{e}^{12}-\left(\frac{16}{15}-\frac{7}{12}{e}^3\right)$

= $-\frac{7}{12}{e}^{12}+\frac{16}{15}-\left(-\frac{7}{12}{e}^3+\frac{16}{15}\right)$

= $-\frac{7}{12}{e}^{12}+\frac{16}{15}+\frac{7}{12}{e}^3-1·\frac{16}{15}$

= $-\frac{7}{12}{e}^{12}+\frac{16}{15}+\frac{7}{12}{e}^3-\frac{16}{15}$

= $-\frac{7}{12}{e}^{12}+\frac{7}{12}{e}^3+\frac{16}{15}-\frac{16}{15}$

= $-\frac{7}{12}{e}^{12}+\frac{7}{12}{e}^3+0$

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

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

= $e^0-e$

= $1-e$

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

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

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

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

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

= $\frac{27}{12}\cdot {\pi }^3-\frac{1}{12}\cdot {\pi }^3$

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

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

$=$ $-\frac{1}{2}{e}^{-2\cdot 4+3}+\frac{1}{2}{e}^{-2\cdot 2+3}$

= $-\frac{1}{2}{e}^{-8+3}+\frac{1}{2}{e}^{-4+3}$

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