Question

Evaluate ∫ ∫D (y/(x5+1)) dA on the region D defined by 0≤x≤1; 0≤y≤x2.

Answer 1

$\iint\limits_{D} \frac{y}{x^5 + 1}dA \\ D = \{0\leq x\leq1, 0\leq y \leq x^2\}\\$

$\iint\limits_{D} \frac{y}{x^5 + 1}dA = \int\limits_{0}^{1}dx\int\limits_{0}^{x^2} \frac{y}{x^5 + 1}dy= \int\limits_{0}^{1}\frac{dx}{x^5+1}\int\limits_{0}^{x^2}ydy = \\ = \int\limits_{0}^{1}\frac{dx}{x^5+1} (\frac{y^2}{2})|_0^{x^2} = \int\limits_{0}^{1}\frac{x^4dx}{2(x^5+1)} = \\ =\left[ \begin{array}{ccc} x^5 +1 = u & 1\to 2\\ 5x^4 dx = du & 0 \to 1\\ \end{array} \right]= \frac{5}{2}\int\limits_{1}^2\frac{du}{u} = \\=\frac{5}{2} ln(u)|_1^2 =\frac{5}{2}(ln2 - ln1) = \frac{5}{2}ln2$