Question Number 197383 by universe last updated on 15/Sep/23
$$\:{evaluate}\:\:\int_{\mathrm{1}/\mathrm{4}} ^{\mathrm{1}} \int_{\sqrt{{x}−{x}^{\mathrm{2}} }} ^{\sqrt{{x}}} \frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }{dydx}\:=\:?? \\ $$
Commented by Frix last updated on 16/Sep/23
![∫_( (√x)(√(1−x))) ^(√x) ((x^2 −y^2 )/x^2 )dy=[y−(y^3 /(3x^2 ))]_((√x)(√(1−x))) ^(√x) = =((3x−1−(4x−1)(√(1−x)))/(3(√x))) ∫_(1/4) ^1 ((3x−1−(4x−1)(√(1−x)))/(3(√x)))dx= =[−((2(√x)(x−1)(1−(√(1−x)))/3)]_(1/4) ^1 =(1/4)−((√3)/8)](https://www.tinkutara.com/question/Q197387.png)
$$\underset{\:\sqrt{{x}}\sqrt{\mathrm{1}−{x}}} {\overset{\sqrt{{x}}} {\int}}\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }{dy}=\left[{y}−\frac{{y}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{2}} }\right]_{\sqrt{{x}}\sqrt{\mathrm{1}−{x}}} ^{\sqrt{{x}}} = \\ $$$$=\frac{\mathrm{3}{x}−\mathrm{1}−\left(\mathrm{4}{x}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}}}{\mathrm{3}\sqrt{{x}}} \\ $$$$\underset{\frac{\mathrm{1}}{\mathrm{4}}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{3}{x}−\mathrm{1}−\left(\mathrm{4}{x}−\mathrm{1}\right)\sqrt{\mathrm{1}−{x}}}{\mathrm{3}\sqrt{{x}}}{dx}= \\ $$$$=\left[−\frac{\mathrm{2}\sqrt{{x}}\left({x}−\mathrm{1}\right)\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}}\right.}{\mathrm{3}}\right]_{\frac{\mathrm{1}}{\mathrm{4}}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$