Evaluate the line integral \$\$int_C (x+2y)dx + x^2dy,\$\$ whereby \$C\$ is composed of line segments from \$(0,0)\$ to \$(2,1)\$ and from \$(2,1)\$ come \$(3,0)\$.

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How do you fix this by using the following parametrics? I break-up them up yet got a an adverse answer of -1/3. What"s wrong?

For \$C_1\$ got, \$langle t, t/2 angle\$, \$0 leq t leq 2\$.

For \$C_2\$ got, \$langle t, 3-y angle\$, \$2 leq t leq 3\$. Hints:

\$\$;;;(0,0) o(2,1),:;;; 0le xle 2;,;;y=frac x2implies\$\$

\$\$intlimits_(0,0)^(2,1)(x+2y)dx+x^2dy=intlimits_0^2 (x+x)dx+x^2left(frac12,dx ight)=intlimits_0^2left(frac12x^2+2x ight)dx=\$\$

\$\$=frac16cdot8+4=frac163\$\$

and something similar with the other one...

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\$\$(2,1) o(3,0):;;;2le xle 3;,;;y=-x+3implies\$\$

\$\$intlimits_(2,1)^(3,0(x+2y)dx+x^2dy=intlimits_2^3 (x+2(-x+3))dx+x^2left((-1),dx ight)=intlimits_2^3left(-x^2-x+6 ight)dx=\$\$

\$\$left.-frac13x^3 ight|_2^3-left.left.frac12x^2 ight|_2^3+6x ight|_2^3=-frac193-frac52+6=-frac176\$\$

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edited Apr 30 "13 in ~ 21:05
answered Apr 30 "13 at 20:56 DonAntonioDonAntonio
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Try writing these integrals as\$\$int_C (x+2y,x^2)cdot (dx,dy)\$\$where \$cdot\$ is the normal inner product.Now, for \$C_1\$, do \$(x,y)=(x(t),y(t))=(t,fract2)\$ (if her parametrization is correct), from which we have actually \$dx=dt\$ and also \$dy=fracdt2\$. You can now easily integrate with respect to \$t\$, in ~ the variety where \$t\$ varies.

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answered Apr 30 "13 in ~ 20:50 MarraMarra
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