Let \(I = \mathop \smallint \nolimits_{x = 0}^1 \mathop \smallint \nolimits_{y = 0}^{{x^2}} x{y^2}dydx.\) Then, I may also be expressed as

\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = 0}^{\sqrt y } y{x^2}dxdy\)

\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = 0}^{\sqrt y } x{y^2}dxdy\)

\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = \sqrt y }^1 y{x^2}dxdy\)

\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = \sqrt y }^1 x{y^2}dxdy\)

Let \(I = \mathop \smallint \nolimits_{x = 0}^1 \mathop \smallint \nol

1.	Let \(I = \mathop \smallint \nolimits_{x = 0}^1 \mathop \smallint \nolimits_{y = 0}^{{x^2}} x{y^2}dydx.\) Then, I may also be expressed as
A.	\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = 0}^{\sqrt y } x{y^2}dxdy\)
B.	\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = \sqrt y }^1 y{x^2}dxdy\)
C.	\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = \sqrt y }^1 x{y^2}dxdy\)
D.	\(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = 0}^{\sqrt y } y{x^2}dxdy\)
Answer» D. \(\mathop \smallint \nolimits_{y = 0}^1 \mathop \smallint \nolimits_{x = 0}^{\sqrt y } y{x^2}dxdy\)

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Let \(I = \mathop \smallint \nolimits_{x = 0}^1 \mathop \smallint \nolimits_{y = 0}^{{x^2}} x{y^2}dydx.\) Then, I may also be expressed as

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