Question Number 203835 by patrice last updated on 29/Jan/24
Answered by MathematicalUser2357 last updated on 30/Jan/24
$${bruh} \\ $$
Commented by Frix last updated on 30/Jan/24
$$\mathrm{It}'\mathrm{s}\:\mathrm{not}\:\mathrm{that}\:\mathrm{hard}… \\ $$
Answered by MrGaster last updated on 03/Nov/24
$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\sqrt{\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} }}}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{6}−\left(\mathrm{6}−\mathrm{5}{x}^{\mathrm{2}} \right)}}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{5}{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \frac{\sqrt{\mathrm{5}}{x}}{\:\sqrt{\mathrm{5}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {x}\:{dx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{0}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}} \\ $$$$=\mathrm{1} \\ $$
Commented by Frix last updated on 03/Nov/24
$$\left(\mathrm{5}{x}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{4}} −\mathrm{60}{x}^{\mathrm{2}} +\mathrm{36}\neq\mathrm{25}{x}^{\mathrm{4}} −\mathrm{50}{x}^{\mathrm{2}} +\mathrm{36} \\ $$