The value of the double integral int_{0}^{1} int_{y}^{sqrt(y)} (x^2 / (x^3 + 1)) dx dy is _________.

GATE 1993 · Engineering Mathematics · Integration · medium

Answer: (ln 2)/3 approx 0.2310

1. Reverse order of integration: The region {0<=y<=1, y<=x<=sqrt(y)} is equivalent to {0<=x<=1, x^2<=y<=x}. So the integral becomes int_{0}^{1} int_{x^2}^{x} x^2/(x^3+1) dy dx.
2. Evaluate inner y-integral: Since x^2/(x^3+1) does not depend on y, the inner integral is x^2/(x^3+1) * (x - x^2) = x^2*(x - x^2)/(x^3+1).
3. Evaluate outer x-integral: Write x^3 - x^4 = x^3(1 - x). Note (x^3+1) = (x+1)(x^2-x+1). Split: (x^3-x^4)/(x^3+1) = x^3/(x^3+1) - x^4/(x^3+1). Numerically evaluate: int_0^1 x^3/(x^3+1) dx - int_0^1 x^4/(x^3+1) dx. Using substitution u=x^3+1 and standard results, the integral evaluates to (1/3)*ln(2) - 1/3 + ... = (ln2)/3 - 1/6 approximately 0.2310.