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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

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{that}\mathrm{hard}...$$

Answered by MrGaster last updated on 03/Nov/24

$$={\int }_0^{\sqrt{2}}\frac{\sqrt{6-\sqrt{{(5x^2-6)}^2}}}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{6-{(5x^2-6)}^2}}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{6-(6-5x^2)}}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{5x^2}}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}\frac{\sqrt{5}x}{\sqrt{5}}dx$$$$={\int }_0^{\sqrt{2}}xdx=\frac{x^2}{2}{\mid }_0^{\sqrt{2}}$$$$=\frac{{\left(\sqrt{2}\right)}^2}{2}-\frac{0^2}{2}$$$$=\frac{2}{2}$$$$=1$$

Commented by Frix last updated on 03/Nov/24

$${(5x^2-6)}^2=25x^4-60x^2+36\ne 25x^4-50x^2+36$$

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