$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $\frac{\mathrm{(5\; -\; 3)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>3</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-6}}{2}$ = -3.

I₃ = $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₃ = (8 - 5) ⋅ ( - 3 ) = 3 ⋅ ( - 3 ) = -9.

I₄ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₄ = (10 - 8) ⋅ $\frac{\mathrm{-3\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>2</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 2.5 ) = -5.

Somit gilt:

$$\underset{3}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₂ + I₃ + I₄ = $$\underset{3}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = -3 -9 -5 = -17

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

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

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

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

= $\frac{8}{6}-\frac{9}{6}-\left(-\frac{8}{6}-\frac{9}{6}\right)$

= $-\frac{1}{6}-1·\left(-\frac{17}{6}\right)$

= $-\frac{1}{6}+\frac{17}{6}$

= $\frac{8}{3}$

= ${\left[-8e^{-x}+\frac{1}{3}{x}^{\frac{3}{2}}\right]}_1^{16}$

= ${\left[-8e^{-x}+\frac{1}{3}{\left(\sqrt{x}\right)}^3\right]}_1^{16}$

$=$ $-8e^{-16}+\frac{1}{3}{\left(\sqrt{16}\right)}^3-\left(-8e^{-1}+\frac{1}{3}{\left(\sqrt{1}\right)}^3\right)$

= $-8e^{-16}+\frac{1}{3}\cdot 4^3-\left(-8e^{-1}+\frac{1}{3}\cdot 1^3\right)$

= $-8e^{-16}+\frac{1}{3}\cdot 64-\left(-8e^{-1}+\frac{1}{3}\cdot 1\right)$

= $-8e^{-16}+\frac{64}{3}-\left(-8e^{-1}+\frac{1}{3}\right)$

= $8e^{-1}-1·\frac{1}{3}-8e^{-16}+\frac{64}{3}$

= $8e^{-1}-\frac{1}{3}-8e^{-16}+\frac{64}{3}$

= $8e^{-1}-8e^{-16}-\frac{1}{3}+\frac{64}{3}$

= $8e^{-1}-8e^{-16}+21$

= ${\left[-\frac{3}{2}{\left(-2x+5\right)}^{-1}\right]}_3^6$

= ${\left[-\frac{3}{2\left(-2x+5\right)}\right]}_3^6$

$=$ $-\frac{3}{2\left(-2\cdot 6+5\right)}+\frac{3}{2\left(-2\cdot 3+5\right)}$

= $-\frac{3}{2\left(-12+5\right)}+\frac{3}{2\left(-6+5\right)}$

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

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

= $\frac{3}{14}-\frac{3}{2}$

= $\frac{3}{14}-\frac{21}{14}$

= $-\frac{9}{7}$

= ${\left[4\cdot \sin \left(x\right)-\frac{7}{3}\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

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

= $4\cdot \left(-1\right)-\frac{7}{3}\cdot 0-\left(4\cdot 1-\frac{7}{3}\cdot 0\right)$

= $-4+0-\left(4+0\right)$

= $-4-4$

= $-8$

= ${\left[\frac{1}{4}{\left(3x-7\right)}^4+2x^2\right]}_2^4$

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

= $\frac{1}{4}\cdot {\left(12-7\right)}^4+2\cdot 16-\left(\frac{1}{4}\cdot {\left(6-7\right)}^4+2\cdot 4\right)$

= $\frac{1}{4}\cdot 5^4+32-\left(\frac{1}{4}\cdot {\left(-1\right)}^4+8\right)$

= $\frac{1}{4}\cdot 625+32-\left(\frac{1}{4}\cdot 1+8\right)$

= $\frac{625}{4}+32-\left(\frac{1}{4}+8\right)$

= $\frac{625}{4}+\frac{128}{4}-\left(\frac{1}{4}+\frac{32}{4}\right)$

= $\frac{753}{4}-1·\frac{33}{4}$

= $\frac{753}{4}-\frac{33}{4}$

= $180$