Question

Evaluate the iterated integral \(\int\limits_0^1 {\int\limits_{2x}^2 {(x - y)dxdy} } \)

Short answer

Value of integral is\(\left( { - 1} \right)\).

Step-by-step solution

Integrating with respect to \(y\) and considering \(x\) as constant.

\(\begin{aligned}\int\limits_0^1 {\int\limits_{2x}^2 {(x - y)dxdy} } &= \int\limits_0^1 {\left( {xy - \frac{{{y^2}}}{2}} \right)_{2x}^2dx} \\ &= \int\limits_0^1 {\left\{ {\left( {2x - \frac{{{2^2}}}{2}} \right) - \left( {x(2x) - \frac{{{{(2x)}^2}}}{2}} \right)} \right\}} dx\\ = \int\limits_0^1 {\left\{ {\left( {2x - \frac{4}{2}} \right) - \left( {2{x^3} - \frac{{4{x^2}}}{2}} \right)} \right\}} dx\\ &= \int\limits_0^1 {\left\{ {\left( {2x - 2} \right) - \left( {2{x^2} - 2{x^2}} \right)} \right\}} dx\\ &= \int\limits_0^1 {\left( {2x - 2} \right)} dx\end{aligned}\)

Finalizing the Solution

\(\begin{array}{c}\int\limits_0^1 {\int\limits_{2x}^2 {(x - y)dxdy} } &= \int\limits_0^1 {(2x - 2)dx} \\ &= \left( {\frac{{2{x^2}}}{2} - 2x} \right)_0^1\\ = ({x^2} - 2x)_0^1\\ = 1 - 2 - 0\\ &= - 1\end{aligned}\)

Hence, the value of integral is\(\left( { - 1} \right)\).