第5題
計算\(\displaystyle \int_0^1 \int_x^1 x^2\sqrt{1+y^4}dydx\)
[解答]
\(\displaystyle \int_0^1 {\int_x^1 {{x^2}\sqrt {1 + {y^4}} dydx} = \int_0^1 {\int_0^y {{x^2}\sqrt {1 + {y^4}} dxdy} } } = \int_0^1 {\left( {\frac{1}{3}{x^3}\sqrt {1 + {y^4}} \left| {_0^y} \right.} \right)dy} \)
\(\displaystyle = \frac{1}{3}\int_0^1 {{y^3}\sqrt {1 + {y^4}} dy} = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{4}{\left( {1 + {y^4}} \right)^{\frac{3}{2}}}\left| {_0^1} \right. = \frac{{\sqrt 2 }}{9} - \frac{1}{{18}} \)