let C ={(x,y)∈R^2 / 0≤x≤1 and y=2x^2 } calculate ∫


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Question Number 36195 by prof Abdo imad last updated on 30/May/18
$let C ={(x,y)∈R^2 / 0≤x≤1 and y=2x^2 } calculate ∫_C x^2 ydx +(x^2 −y^2 )dy$
$l e t C = {(x, y) \in R^{2} / 0 ⩽ x ⩽ 1 a n d y = 2 x^{2}}$ $c a l c u l a t e \int_{C} x^{2} y d x + (x^{2} - y^{2}) d y$
Commented by math khazana by abdo last updated on 18/Aug/18
$∫_C x^2 ydx +(x^2 −y^2 )dy =∫_C x^2 (2x^2 )dx +∫_C (x^2 −4x^4 )(4x)dx =∫_0 ^1 4x^4 dx + 4∫_0 ^1 (x^3 −4x^5 )dx =(4/5)[x^5 ]_0 ^1 + 4 [ (x^4 /4) −(2/3) x^6 ]_0 ^1 =(4/5) +4{(1/4) −(2/3)} =(4/5) +1 −(8/3) = (9/5) −(8/3) =((27−40)/(15)) =−((13)/(15)) .$
$\int_{C} x^{2} y d x + (x^{2} - y^{2}) d y$ $= \int_{C} x^{2} (2 x^{2}) d x + \int_{C} (x^{2} - 4 x^{4}) (4 x) d x$ $= \int_{0}^{1} 4 x^{4} d x + 4 \int_{0}^{1} (x^{3} - 4 x^{5}) d x$ $= \frac{4}{5} {[x^{5}]}_{0}^{1} + 4 {[\frac{x^{4}}{4} - \frac{2}{3} x^{6}]}_{0}^{1} = \frac{4}{5} + 4 {\frac{1}{4} - \frac{2}{3}}$ $= \frac{4}{5} + 1 - \frac{8}{3} = \frac{9}{5} - \frac{8}{3} = \frac{27 - 40}{15} = - \frac{13}{15} .$
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