## anonymous 3 years ago Evaluate the line integral 5ydx+2xdy where C is the straight line path from (4,4) to (7,8)

1. anonymous

I got 276 but was wrong

2. anonymous

3. anonymous

$\int_C (5y~dx+2x~dy)$ The line connecting (4,4) and (7,8) is $$y=\dfrac{4}{3}x-\dfrac{4}{3}$$ Parameterize $$C$$ by $\begin{cases}x=t\\ \\ y=\dfrac{4}{3}t-\dfrac{4}{3}\\ \\ 4<t<7 \end{cases}$ So you have $$dx=dt$$ and $$dy=\dfrac{4}{3}dt$$: $\int_4^7 \bigg[5\left(\dfrac{4}{3}t-\dfrac{4}{3}\right)~dt+2t~\left(\dfrac{4}{3}dt\right)\bigg]\\ \int_4^7\bigg[\frac{28}{3}t-\frac{20}{3}\bigg]~dt\\ ~~~~~~~~~~~~~~~~\vdots$

4. anonymous

did you get 134?

5. anonymous

Yep: http://www.wolframalpha.com/input/?i=Integrate%5B%2828%2F3%29*t-%2820%2F3%29%2C%7Bt%2C4%2C7%7D%5D Can't guarantee that's the right answer, but that's what my method is getting, so I'm sticking with it.

6. anonymous

when i did it by hand, i ended up getting 288

7. anonymous

Could you show your work? It might be a mistake in the integration, or the parameterization.

8. anonymous

288 is wrong :(

9. anonymous

Well what did you do when you tried it? How did you parameterize $$C$$?

10. anonymous

no i integrated wrong! the answer is indeed 134

11. anonymous

Oh ok then.

12. anonymous

Thanks so much

13. anonymous

You're welcome!