Evaluate the line integral where

$$ \int_C ydx + x^2dy $$

  • C1 is the path of the straight line segment from the origin, (0,0) to the point (2,18)
  • C2 is the path of the parabola $y=−x^2+8x+6$ from the point (2,18) to the point (5,21)

First I looked at just c1. I found that y=9x. Then I took the integral of $$ \int_0^2 (9x+9x^2)dx =42 $$

Then I looked at c2 where I took the integral $$ \int_2^5[(-x^2+8x+6) + x^2(-2x+8)]dx $$ $$ \int_2^5(-2x^3+7x^2+8x+6)dx = 58.5 $$

I added what I got from C1 and C2 to get 100.5 but this is not correct. Can someone look at my work and see where I am going wrong?

  • 1
    $\begingroup$ I think the second integral is wrong. I get $70.5$ evaluating it. If the correct answer in the end is $112.5$ this might be it. $\endgroup$ – mickep Apr 4 '15 at 20:26
  • $\begingroup$ @mickep That is correct. What did I mess up in my second integral? $\endgroup$ – Ayoshna Apr 4 '15 at 20:29
  • $\begingroup$ No idea, since the left-hand side is correct, and the right-hand side is not, and you do not provide any further details. $\endgroup$ – mickep Apr 4 '15 at 20:30

I've solved the two integral by two distinct methods,


$$\int_{C_2}=\int_{2}^5(-x^2+8x+6)dx+\int_2^5x^2(-2x+8)dx$$ $$=\int_2^{5}7x^2+8x+6-2x^3dx=383-\frac{625}{2}.$$

Putting $\int_C Fdr=\int_{C_{1}}Fdr+\int_{C_{2}}Fdr$ for $F(x,y)=(y,x^2)$, then $F(x(t),y(t))=(y(t),(x(t))^2)$.

Thus on $C_1$ we have: $F(t)=F(x(t),y(t))=(9t,(t)^2)$ since $r(t)=(x(t),y(t))=(t,9t)$.

And on $C_2$ we have: $F(t)=F(x(t),y(t))=(-t^2+8t+6,(t)^2)$ since $r(t)=(x(t),y(t))=(t,-t^2+8t+6)$.

And limits in integrals are the same because $x(t)$ is equals to $t$ in both cases. In your book you can find the following method and compute as follows since the functions and paths satisfy the requiered hypothesis:

$$\int_{C_{1}}Fdr=\int_{0}^2 F(x(t),y(t))\cdot r'(t)dt=\int_0^2(9t,t^2)\cdot (1,9)dt=\int_0^2 9t+9t^2dt=42.$$

$$\int_{C_{2}}Fdr=\int_{2}^5 F(x(t),y(t))\cdot r'(t)dt=\int_2^5(-t^2+8t+6,t^2)\cdot (1,-2t+8)dt=$$ $$=\int_2^5 -2t^3+7t^2+8t+6dt=\left[-\frac{2t^4}{4}+\frac{7t^3}{3}+\frac{8t^2}{2}+6t\right]_{t=2}^{t=5}=$$


  • $\begingroup$ Very thanks much $\endgroup$ – user243301 Sep 9 '15 at 5:13

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