Question

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Answer 1

$intlimits^4_{-2} {3sqrt{3-frac{x}{3}}-frac{81}{(2x-5)^3}} , dx=\ intlimits^4_{-2} {3sqrt{3-frac{x}{3}}} , dx-intlimits^4_{-2} {frac{81}{(2x-5)^3}} , dx=\ 9intlimits^4_{-2} {sqrt{3-frac{x}{3}}} , d(3-frac{x}{3})-frac{81}{2}intlimits^4_{-2} {(2x-5)^{-3}} , d(2x-5)=\ 9*frac{(3-frac{x}{3})^{frac{3}{2}}}{frac{3}{2}}|^4_{-2}-frac{81}{2}frac{{(2x-5)^{-2}}}{-2}|^4_{-2}=\ 18(3-frac{x}{3})^{frac{3}{2}}|^4_{-2}+frac{81(2x-5)^{-2}}{4}|^4_{-2}=$

$18*(3-frac{4}{3})^{frac{3}{2}}-18*(3-frac{-2}{3})^{frac{3}{2}}+frac{81(2*4-5)^{-2}}{4}-frac{81(2*(-2)-5)^{-2}}{4}=\ 18*frac{5}{3}*sqrt {frac{5}{3}}-18*frac{11}{3}sqrt{frac{11}{3}}+frac{9}{4}-frac{1}{4}=\ 30*sqrt {frac{5}{3}}-66*sqrt{frac{11}{3}}+2$

 

 

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