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Question Number 173062 by Gbenga last updated on 06/Jul/22

Commented by kaivan.ahmadi last updated on 06/Jul/22
$1. D={−2≤x≤2,4x−x^2 ≤y≤4} ∫_(−2) ^2 ∫_(4x−x^2 ) ^4 dydx=∫_(−2) ^2 (4−4x+x^2 )dx= (4x−2x^2 +(x^3 /3))∣_(−2) ^2 = (8−8+(8/3))−(−8−8−(8/3))=16+((16)/3)=((64)/3)$
$1 .$ $D = {- 2 ⩽ x ⩽ 2, 4 x - x^{2} ⩽ y ⩽ 4}$ $\int_{- 2}^{2} \int_{4 x - x^{2}}^{4} d y d x = \int_{- 2}^{2} (4 - 4 x + x^{2}) d x =$ $(4 x - 2 x^{2} + \frac{x^{3}}{3}) ∣_{- 2}^{2} =$ $(8 - 8 + \frac{8}{3}) - (- 8 - 8 - \frac{8}{3}) = 16 + \frac{16}{3} = \frac{64}{3}$
Commented by Gbenga last updated on 06/Jul/22

$great work sir$
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