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W warwick Full Member Joined Jan 27, 2006 Messages 311 Apr 23, 2008 #1 See Image. Attachments review.jpg 190.8 KB · Views: 148
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Apr 23, 2008 #2 ∫(2x−y2)dx+(y3−2xy)dy\displaystyle \int{(2x-y^{2})dx+(y^{3}-2xy)dy}∫(2x−y2)dx+(y3−2xy)dy (1,1) to (3,2) I can't see all your picture, but you have: x=1+2t, y=1+t, 0≤t≤1\displaystyle x=1+2t, \;\ y=1+t, \;\ 0\leq{t}\leq{1}x=1+2t, y=1+t, 0≤t≤1 ∫[(2x−y2)i+(y3−2xy)j]⋅(dxi+dyj)\displaystyle \int\left[(2x-y^{2})i+(y^{3}-2xy)j\right]\cdot(dxi+dyj)∫[(2x−y2)i+(y3−2xy)j]⋅(dxi+dyj) ∫01(t−1)(t2−2t−1)dt=34\displaystyle \int_{0}^{1}(t-1)(t^{2}-2t-1)dt=\frac{3}{4}∫01(t−1)(t2−2t−1)dt=43
∫(2x−y2)dx+(y3−2xy)dy\displaystyle \int{(2x-y^{2})dx+(y^{3}-2xy)dy}∫(2x−y2)dx+(y3−2xy)dy (1,1) to (3,2) I can't see all your picture, but you have: x=1+2t, y=1+t, 0≤t≤1\displaystyle x=1+2t, \;\ y=1+t, \;\ 0\leq{t}\leq{1}x=1+2t, y=1+t, 0≤t≤1 ∫[(2x−y2)i+(y3−2xy)j]⋅(dxi+dyj)\displaystyle \int\left[(2x-y^{2})i+(y^{3}-2xy)j\right]\cdot(dxi+dyj)∫[(2x−y2)i+(y3−2xy)j]⋅(dxi+dyj) ∫01(t−1)(t2−2t−1)dt=34\displaystyle \int_{0}^{1}(t-1)(t^{2}-2t-1)dt=\frac{3}{4}∫01(t−1)(t2−2t−1)dt=43
W warwick Full Member Joined Jan 27, 2006 Messages 311 Apr 23, 2008 #3 Yeah, that's the answer I got. The back of the review has -27/4.
