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Question-201762




Question Number 201762 by Calculusboy last updated on 11/Dec/23
Answered by mr W last updated on 12/Dec/23
∫_(−2) ^2 (x^3 cos (x/2)+(1/2))(√(4−x^2 )) dx  =∫_(−2) ^2 (x^3 cos (x/2))(√(4−x^2 )) _(odd function) dx+∫_(−2) ^2 (1/2)(√(4−x^2 )) _(even function) dx  =0+∫_0 ^2 (√(4−x^2 )) dx  =((π×2^2 )/4)=π ✓
$$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} \mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\int_{−\mathrm{2}} ^{\mathrm{2}} \underset{{odd}\:{function}} {\left({x}^{\mathrm{3}} \mathrm{cos}\:\frac{{x}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:}{dx}+\int_{−\mathrm{2}} ^{\mathrm{2}} \underset{{even}\:{function}} {\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:}{dx} \\ $$$$=\mathrm{0}+\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{\pi×\mathrm{2}^{\mathrm{2}} }{\mathrm{4}}=\pi\:\checkmark \\ $$
Commented by Calculusboy last updated on 11/Dec/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by mr W last updated on 12/Dec/23
if f(−x)=−f(x), then  ∫_(−a) ^a f(x)dx=0.  if f(−x)=f(x), then  ∫_(−a) ^a f(x)dx=2∫_0 ^a f(x)dx
$${if}\:{f}\left(−{x}\right)=−{f}\left({x}\right),\:{then} \\ $$$$\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{0}. \\ $$$${if}\:{f}\left(−{x}\right)={f}\left({x}\right),\:{then} \\ $$$$\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx} \\ $$

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