Question

Evaluate \(\iint\limits_D {\frac{y}{{1 + {x^5}}}dA,D = \{ (x,y)/0 \leqslant x \leqslant 1,0 \leqslant y \leqslant {x^2}\} }\)

Short answer

value is \(\frac{{ln(2)}}{{10}}\)

Step-by-step solution

Step 1:- Integrate with respect to y

\(I = \int\limits_0^1 {\frac{1}{{{x^5} + 1}}\left( {\frac{1}{2}{y^2}} \right)} \int\limits_0^{{x^2}} {dx} \)

\( = \frac{1}{2}\int\limits_0^1 {\frac{1}{{{x^5} + 1}}({x^4} - 0)dx} \)

\( = \frac{1}{2}\int\limits_0^1 {\frac{{{x^4}}}{{{x^5} + 1}}dx} \)

Step 2:- Integrate on x 

\(\begin{array}{l}I = \frac{1}{2}\int\limits_1^2 {\frac{1}{5}\left( {\frac{1}{u}} \right)} du.\\ = \frac{1}{{10}}\left( {ln2 - ln1} \right)_1^2\\ = \frac{1}{{10}}\left( {ln2 - ln1} \right)\\ = \frac{1}{{10}}ln2.\end{array}\)

Therefore, value is \(\frac{{ln2}}{{10}}\)