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SSC Pure Mathematics QUESTION #1267
Question 1
Changing the order of integration in \(\displaystyle\int_0^2\int_0^{\sqrt{x}}f(x,y)\,dy\,dx\) gives:
  • \(\displaystyle\int_0^{\sqrt{2}}\int_0^{y^2}f(x,y)\,dx\,dy\)
  • \(\displaystyle\int_0^{2}\int_0^{y^2}f(x,y)\,dx\,dy\)
  • \(\displaystyle\int_0^{\sqrt{2}}\int_{y^2}^{2}f(x,y)\,dx\,dy\)✔️
  • \(\displaystyle\int_0^{2}\int_{\sqrt{x}}^{2}f(x,y)\,dx\,dy\)
Correct Answer Explanation
Original region: \(0\leq x\leq 2\), \(0\leq y\leq\sqrt{x}\). Equivalently: \(y\) ranges from \(0\) to \(\sqrt{2}\); for fixed \(y\), \(x\) ranges from \(y^2\) to \(2\) (since \(y\leq\sqrt{x}\Rightarrow x\geq y^2\)). So: \(\int_0^{\sqrt{2}}\int_{y^2}^{2}f(x,y)\,dx\,dy\).