$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₁ = (2 - 0) ⋅ 4 = 2 ⋅ 4 = 8.

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 = $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(6\; -\; 4)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-8}}{2}$ = -4.

I₄ = $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ : Rechtecksfläche I₄ = (8 - 6) ⋅ ( - 4 ) = 2 ⋅ ( - 4 ) = -8.

I₅ = $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ : Trapezfläche I₅ = (10 - 8) ⋅ $\frac{\mathrm{-4\; +\; <mo>(</mo>\; <mo>-</mo>\; <mn>5</mn>\; <mo>)</mo>}}{2}$ = 2 ⋅ ( - 4.5 ) = -9.

Somit gilt:

$$\underset{0}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ + I₄ + I₅ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{4}{\int }}\mathrm{f(x)}dx$$ + $$\underset{4}{\overset{6}{\int }}\mathrm{f(x)}dx$$ + $$\underset{6}{\overset{8}{\int }}\mathrm{f(x)}dx$$ + $$\underset{8}{\overset{10}{\int }}\mathrm{f(x)}dx$$ = 8 +4 -4 -8 -9 = -9

= ${\left[-x^2-x\right]}_1^5$

$=$ $-5^2-5-\left(-1^2-1\right)$

= $-25-5-\left(-1-1\right)$

= $-25-5-1·\left(-1\right)-1·\left(-1\right)$

= $-25-5+1+1$

= $-28$

= ${\left[-\frac{7}{4}\cdot \sin \left(x\right)-\frac{5}{3}{e}^{3x}\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $-\frac{7}{4}\cdot \sin \left(\frac{3}{2}\pi \right)-\frac{5}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(-\frac{7}{4}\cdot \sin \left(\frac{1}{2}\pi \right)-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{7}{4}\cdot \left(-1\right)-\frac{5}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(-\frac{7}{4}\cdot 1-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $\frac{7}{4}-\frac{5}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}-\left(-\frac{7}{4}-\frac{5}{3}{e}^{3\cdot \left(\frac{1}{2}\pi \right)}\right)$

= $-\frac{5}{3}{e}^{3\cdot \left(\frac{3}{2}\pi \right)}+\frac{7}{4}-\left(-\frac{5}{3}{e}^{\frac{3}{2}\pi }-\frac{7}{4}\right)$

= $-\frac{5}{3}{e}^{\frac{9}{2}\pi }+\frac{7}{4}+\frac{5}{3}{e}^{\frac{3}{2}\pi }-1·\left(-\frac{7}{4}\right)$

= $-\frac{5}{3}{e}^{\frac{9}{2}\pi }+\frac{7}{4}+\frac{5}{3}{e}^{\frac{3}{2}\pi }+\frac{7}{4}$

= $-\frac{5}{3}{e}^{\frac{9}{2}\pi }+\frac{5}{3}{e}^{\frac{3}{2}\pi }+\frac{7}{4}+\frac{7}{4}$

= $-\frac{5}{3}{e}^{\frac{9}{2}\pi }+\frac{5}{3}{e}^{\frac{3}{2}\pi }+\frac{7}{2}$

= ${\left[e^{-3x+6}\right]}_2^4$

$=$ $e^{-3\cdot 4+6}-e^{-3\cdot 2+6}$

= $e^{-12+6}-e^{-6+6}$

= $e^{-6}-e^0$

= $e^{-6}-1$

= ${\left[-\frac{1}{2}\cdot \cos \left(x\right)-5\cdot \sin \left(x\right)\right]}_0^{\frac{1}{2}\pi }$

$=$ $-\frac{1}{2}\cdot \cos \left(\frac{1}{2}\pi \right)-5\cdot \sin \left(\frac{1}{2}\pi \right)-\left(-\frac{1}{2}\cdot \cos \left(0\right)-5\cdot \sin \left(0\right)\right)$

= $-\frac{1}{2}\cdot 0-5\cdot 1-\left(-\frac{1}{2}\cdot 1-5\cdot 0\right)$

= $0-5-\left(-\frac{1}{2}+0\right)$

= $-5-\left(-\frac{1}{2}+0\right)$

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

= $-\frac{10}{2}+\frac{1}{2}$

= $-\frac{9}{2}$

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

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

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

= $2\cdot 0-2\cdot \left(-1\right)$

= $0+2$

= $2$