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if-f-x-x-16-x-2-3-find-0-5-3-5-f-x-dx-using-trapezoidal-method-then-find-the-max-and-min-value-of-the-error-with-the-given-n-steps-x-n-f-x-n-




Question Number 193036 by ali009 last updated on 02/Jun/23
if f(x)=x(√((16−x^2 )^3 ))  find ∫_(0.5) ^(3.5) f(x) dx using trapezoidal method  then find the max and min value of the error   with the given n steps  x_n                          f(x_n )  −−                   −−−  0.5                        31.24  0.93                      54.69  1.36                     72.3  1.79                     81.98  2.21                    81.25  2.64                    71.54  3.07                    51.68  3.5                      25.41
$${if}\:{f}\left({x}\right)={x}\sqrt{\left(\mathrm{16}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$${find}\:\int_{\mathrm{0}.\mathrm{5}} ^{\mathrm{3}.\mathrm{5}} {f}\left({x}\right)\:{dx}\:{using}\:{trapezoidal}\:{method} \\ $$$${then}\:{find}\:{the}\:{max}\:{and}\:{min}\:{value}\:{of}\:{the}\:{error} \\ $$$$\:{with}\:{the}\:{given}\:{n}\:{steps} \\ $$$${x}_{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}_{{n}} \right) \\ $$$$−−\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−− \\ $$$$\mathrm{0}.\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{31}.\mathrm{24} \\ $$$$\mathrm{0}.\mathrm{93}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{54}.\mathrm{69} \\ $$$$\mathrm{1}.\mathrm{36}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{72}.\mathrm{3} \\ $$$$\mathrm{1}.\mathrm{79}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{81}.\mathrm{98} \\ $$$$\mathrm{2}.\mathrm{21}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{81}.\mathrm{25} \\ $$$$\mathrm{2}.\mathrm{64}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{71}.\mathrm{54} \\ $$$$\mathrm{3}.\mathrm{07}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{51}.\mathrm{68} \\ $$$$\mathrm{3}.\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{25}.\mathrm{41} \\ $$$$ \\ $$
Answered by MM42 last updated on 02/Jun/23
u_f =Σu_i f×Δ_i x=(54.69+72.3+81.98+81.25+71.54+51.68)×0.43+81.98×0.42  =212.2108  l_f =Σl_i f×Δ_i x=(31.24+54.69+72.3+81.25+71.54+51.68+25.41)×0.43+81.25×0.42  =211.9042
$${u}_{{f}} =\Sigma{u}_{{i}} {f}×\Delta_{{i}} {x}=\left(\mathrm{54}.\mathrm{69}+\mathrm{72}.\mathrm{3}+\mathrm{81}.\mathrm{98}+\mathrm{81}.\mathrm{25}+\mathrm{71}.\mathrm{54}+\mathrm{51}.\mathrm{68}\right)×\mathrm{0}.\mathrm{43}+\mathrm{81}.\mathrm{98}×\mathrm{0}.\mathrm{42} \\ $$$$=\mathrm{212}.\mathrm{2108} \\ $$$${l}_{{f}} =\Sigma{l}_{{i}} {f}×\Delta_{{i}} {x}=\left(\mathrm{31}.\mathrm{24}+\mathrm{54}.\mathrm{69}+\mathrm{72}.\mathrm{3}+\mathrm{81}.\mathrm{25}+\mathrm{71}.\mathrm{54}+\mathrm{51}.\mathrm{68}+\mathrm{25}.\mathrm{41}\right)×\mathrm{0}.\mathrm{43}+\mathrm{81}.\mathrm{25}×\mathrm{0}.\mathrm{42} \\ $$$$=\mathrm{211}.\mathrm{9042} \\ $$

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