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Question Number 147904 by puissant last updated on 24/Jul/21

Answered by Olaf_Thorendsen last updated on 24/Jul/21

$$\mathrm{Il}\mathrm{y}\mathrm{a}6\mathrm{valeurs}\mathrm{de}x\left(x_0\mathrm{a}{x}_5\right),\mathrm{ce}\mathrm{qui}$$$$\mathrm{definit}5\mathrm{intervalles}\mathrm{qui}\mathrm{correspondent}$$$$\mathrm{a}5\mathrm{trapezes}.$$$$\mathrm{Chaque}\mathrm{trapeze}k\mathrm{a}\mathrm{une}\mathrm{surface}\mathrm{egale}$$$$\mathrm{a}\frac{y_k+y_{k+1}}{2}(x_{k+1}-x_k).$$$$$$$$\mathrm{La}\mathrm{valeur}\mathrm{approchee}\mathrm{de}{\mathrm{l}}^{\prime }\mathrm{integrale}\mathrm{est}$$$$\mathrm{donc}\underset{k=0}{\sum ^5}\frac{y_k+y_{k+1}}{2}(x_{k+1}-x_k).$$$$\mathrm{Mais}\mathrm{les}\mathrm{longueurs}{\mathrm{d}}^{\prime }\mathrm{intervalles}\mathrm{sont}$$$$\mathrm{constantes}\mathrm{et}\mathrm{egales}\mathrm{a}3:x_{k+1}-x_k=3.$$$$\mathrm{Et}\mathrm{donc}:$$$${\int }_6^{21}ydx\approx \frac{3}{2}\underset{k=0}{\sum ^5}(y_k+y_{k+1})\mathrm{soit}:$$$$\frac{3}{2}(y_0+2y_1+2y_2+2y_3+2y_4+2y_5+y_6)$$$$$$$$\mathrm{Application}\mathrm{numerique}:$$$$\frac{3}{2}(0,3+2\times 0,8+2\times 1,4+2\times 2,1$$$$+2\times 3,0+4,3)=\frac{144}{5}=28,8$$

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