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The value of the integral is:

Aakash , 10 Years ago
Grade 12th pass
anser 1 Answers
jagdish singh singh
\hspace{-0.6 cm }$Let $\bf{I = \int_{2}^{4}\frac{\sqrt{x^2-4}}{x^4}dx\;,}$ Now Put $\bf{x=\frac{1}{t}}$ and $\bf{dx = -\frac{1}{t^2}dt}$ and changing\\\\ \\limits, We get $\bf{I = \int_{\frac{1}{4}}^{\frac{1}{2}}t\sqrt{1-4t^2}dt\;,}$ Now Put $\bf{(1-4t^2) = v\;, }$ Then $\bf{-8tdt = dv}$\\\\\\ And changing limit, So we get $\bf{I = \frac{1}{8}\int_{0}^{\frac{3}{4}}\sqrt{v}dv = = \frac{1}{12}\left[v\sqrt{v}\right]_{0}^{\frac{3}{4}} = \frac{3}{32}.}$
Last Activity: 10 Years ago
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