Menu Close

find-g-x-0-x-e-t-1-t-2-dt-




Question Number 36434 by prof Abdo imad last updated on 02/Jun/18
find g(x) =∫_0 ^x    (e^(−t) /( (√(1+t^2 ))))dt.
$${find}\:{g}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{e}^{−{t}} }{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
((dg(x))/dx)=∫_0 ^x (∂/∂x)((e^(−t) /( (√(1+t^2  )))))dt+(dx/dx)×(e^(−x) /( (√(1+x^2 ))))−  ((dg(x))/dx)=(e^(−x) /( (√(1+x^2 ))))
$$\frac{{dg}\left({x}\right)}{{dx}}=\int_{\mathrm{0}} ^{{x}} \frac{\partial}{\partial{x}}\left(\frac{{e}^{−{t}} }{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} \:}}\right){dt}+\frac{{dx}}{{dx}}×\frac{{e}^{−{x}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}− \\ $$$$\frac{{dg}\left({x}\right)}{{dx}}=\frac{{e}^{−{x}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *