Question

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

Answer 1

I got 276 but was wrong

Answer 2

@Hero please help!

Answer 3

\[\int_C (5y~dx+2x~dy)\]
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 \(dy=\dfrac{4}{3}dt\):
\[\int_4^7 \bigg[5\left(\dfrac{4}{3}t-\dfrac{4}{3}\right)~dt+2t~\left(\dfrac{4}{3}dt\right)\bigg]\\
\int_4^7\bigg[\frac{28}{3}t-\frac{20}{3}\bigg]~dt\\
~~~~~~~~~~~~~~~~\vdots\]

Answer 4

did you get 134?

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

288 is wrong :(

Answer 9

Well what did you do when you tried it? How did you parameterize \(C\)?

Answer 10

no i integrated wrong! the answer is indeed 134

Answer 11

Oh ok then.

Answer 12

Thanks so much

Answer 13

You're welcome!