$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ 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 = $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₂ = $\frac{\mathrm{(5\; -\; 2)\; \cdot \; <mo>(</mo>\; <mo>-</mo>\; <mn>4</mn>\; <mo>)</mo>}}{2}$ = $\frac{\mathrm{-12}}{2}$ = -6.

I₃ = $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ : Dreiecksfläche I₃ = $\frac{\mathrm{(7\; -\; 5)\; \cdot \; <mn>3</mn>}}{2}$ = $\frac{6}{2}$ = 3.

Somit gilt:

$$\underset{0}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = I₁ + I₂ + I₃ = $$\underset{0}{\overset{2}{\int }}\mathrm{f(x)}dx$$ + $$\underset{2}{\overset{5}{\int }}\mathrm{f(x)}dx$$ + $$\underset{5}{\overset{7}{\int }}\mathrm{f(x)}dx$$ = -8 -6 +3 = -11

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

$=$ $2^2+5\cdot 2-\left({\left(-2\right)}^2+5\cdot \left(-2\right)\right)$

= $4+10-\left(4-10\right)$

= $4+10-1·4-1·\left(-10\right)$

= $4+10-4+10$

= $20$

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

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

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

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

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

= $-\frac{1}{6\pi }+\frac{1}{3\pi }$

= $-\frac{1}{6\pi }+\frac{2}{6\pi }$

= $\frac{1}{6\pi }$

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

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

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

= $-\frac{2}{3}\cdot \left(-64\right)+\frac{2}{3}\cdot \left(-1\right)$

= $\frac{128}{3}-\frac{2}{3}$

= $42$

= ${\left[\sin \left(x\right)+5\cdot \cos \left(x\right)\right]}_{\frac{1}{2}\pi }^{\frac{3}{2}\pi }$

$=$ $\sin \left(\frac{3}{2}\pi \right)+5\cdot \cos \left(\frac{3}{2}\pi \right)-\left(\sin \left(\frac{1}{2}\pi \right)+5\cdot \cos \left(\frac{1}{2}\pi \right)\right)$

= $-1+5\cdot 0-\left(1+5\cdot 0\right)$

= $-1+0-\left(1+0\right)$

= $-1-1$

= $-2$

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

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

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

= $-\frac{2}{3}\cdot 0+\frac{2}{3}\cdot 1$

= $0+\frac{2}{3}$

= $0+\frac{2}{3}$

= $\frac{2}{3}$