silemus.

49Double integral, inner yy first

Evaluate ∫01∫0xxy dy dx\displaystyle \int_0^1 \int_0^x xy\,dy\,dx.

Do the inner integral in y first, treating x as constant: ∫₀ˣ xy dy = x³/2. Then integrate x³/2 from 0 to 1 using the power rule to get x⁴/8, which evaluates to 1/8. Switching the order of integration confirms the same answer, 1/8.

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