Question

\(\int\limits_0^1 {\int\limits_x^1 {{e^{{x \mathord{\left/{\vphantom {x y}} \right.} y}}}dydx} }\)

Short answer

Thus, the value of integral is\(\frac{{e - 1}}{2}\).

Step-by-step solution

Given Data.

Consider the integral:

\(\int\limits_0^1 {\int\limits_x^1 {{e^{{x \mathord{\left/ {\vphantom {x y}} \right.} y}}}dydx} } \)

The objective is to reverse the integration and evaluate the integral.

Reversing Order of Integration.

\(\begin{aligned}\int\limits_0^1 {\int\limits_x^1 {{e^{{x \mathord{\left/{\vphantom {x y}} \right.} y}}}dydx} } &= \int\limits_0^1 {\int\limits_0^y {{e^{{x \mathord{\left/{\vphantom {x y}} \right.} y}}}dx} } dy\\ &= \int\limits_0^1 {\left( {\int\limits_0^y {{e^{{x \mathord{\left/{\vphantom {x y}} \right.} y}}}dx} } \right)} dy\\ &= \int\limits_0^1 {\left( {y{e^{{x \mathord{\left/{\vphantom {x y}} \right.} y}}}} \right)_0^y} dy\\ &= \int\limits_0^1 {\left( {ye - y\left( 1 \right)} \right)} dy\\ &= \left( {e - 1} \right)\int\limits_0^1 y dy\\ &= \left( {e - 1} \right)\left( {\frac{{{y^2}}}{2}} \right)_0^1\\ &= \left( {e - 1} \right)\left( {\frac{{{1^2}}}{2} - 0} \right)\\ &= \frac{{e - 1}}{2}\end{aligned}\)

Therefore, the value of integral is\(\frac{{e - 1}}{2}\).