Question

помогите..............................

Answer 1

$intlimits^frac{p}{4}_0 {sin 2x cos 2x} , dx = -frac{1}{2}cos 2x frac{1}{2} sin 2x =-frac{1}{2} cos frac{P}{2} frac{1}{2} sin frac{P}{2} -(-frac{1}{2}cos 0 frac{1}{2}sin 0)=0.$

Answer 2

$intlimits^2_{-1} {((1-x)^4)-frac{1}{sqrt{5x+6}}} , dx=\ intlimits^2_{-1} {(1-x)^4} , dx-intlimits^2_{-1} {frac{1}{sqrt{5x+6}}} , dx=\ -intlimits^2_{-1} {(1-x)^4} , d(1-x)-frac{1}{5}intlimits^2_{-1} {frac{1}{sqrt{5x+6}}} , d(5x+6)=\ -frac{(1-x)^5}{5}|^2_{-1}-frac{1}{5}*frac{sqrt{5x+6}}{0.5}|^2_{-1}=\ -frac{(1-2)^5}{5}+frac{(1-(-1))^5}{5}-frac{1}{5}*frac{sqrt{5*2+6}}{0.5}+frac{1}{5}*frac{sqrt{5*(-1)+6}}{0.5}=\ frac{1}{5}+frac{32}{5}-frac{8}{5}+frac{2}{5}=6.6-1.2=5.4$

 

$intlimits^{frac{pi}{4}}_0 {sin(2x)cos(2x)} , dx=\ frac{1}{2}intlimits^{frac{pi}{4}}_0 {2sin(2x)cos(2x)} , dx=\ frac{1}{2}intlimits^{frac{pi}{4}}_0 {sin(2*2x)} , dx=\ frac{1}{2}intlimits^{frac{pi}{4}}_0 {sin(4x)} , dx=\ frac{1}{8}intlimits^{frac{pi}{4}}_0 {sin(4x)} , d(4x)=\ frac{1}{8}*(-cos(4x))|^{frac{pi}{4}}_0=\ frac{1}{8}*(-cos(4*frac{pi}{4}))-frac{1}{8}*(-cos(4*0))=\ 0.125+0.125=0.25$