# Math Help - Line Integral

1. ## Line Integral

If $F(x)= (3x²y, 2x-y)$, find

$
\int\limits_\gamma F(x)\,dx
$

where $\gamma$ is the directed path from (0,0) to (1,1) along the graph of the vector equation:

$x = (\sin t, 2t/\pi), (0 \leq t \leq \pi/2)$

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Here's what I did:

I recognized that the integral is not independent of path... So set up:

$\int M dx + N dy$

which gives me the sum of 4 integral terms... I evaluated these up to the following step so somebody could double check:

$6/\pi [\pi/2 -1] - 6/\pi[\pi/3 - 7/9] + 4/\pi - 1/2$

Now does this answer seem correct to you?

2. I do NOT get the sum of four terms, I get the sum of three terms:

On the path x= sin t, $y= 2t/\pi$, dx= cos t dt and $dy= 2/\pi dt$.

$3x^2y= 6 tsin^2(t)/\pi$ and $2x- y= 2sin(t)- 2t/\pi$

The integral becomes $6 \int_{t= 0}^{\pi/2} t sin^2 t cos(t) \, dt+ 4\pi\int_{t=0}^{\pi/2}sin(t) \, dt- 4\pi\int_{t=0}^{\pi/2} t\, dt$.

Use integration by parts to do the first integral.

3. Thank you HallsofIvy

Originally Posted by HallsofIvy
The integral becomes $6 nt_{t= 0}^{\pi/2} t sin^2 t cos(t)dt+ 4\pi\int_{t=0}^{\pi/2}sin(t)dt- 4\pi\int_{t=0}^{\pi/2} t dt$.

Use integration by parts to do the first integral.
But I think your integral is incorrect. Did you mean:

$6/\pi \int_{t= 0}^{\pi/2} t sin^2 t cos(t)dt+ 4/\pi\int_{t=0}^{\pi/2}sin(t)dt- 4/\pi^2\int_{t=0}^{\pi/2} t dt$

If this is correct, how would you integrate the first one by parts?

4. Originally Posted by keysar7
Thank you HallsofIvy

But I think your integral is incorrect. Did you mean:

$6/\pi \int_{t= 0}^{\pi/2} t sin^2 t cos(t)dt+ 4/\pi\int_{t=0}^{\pi/2}sin(t)dt- 4/\pi^2\int_{t=0}^{\pi/2} t dt$

If this is correct, how would you integrate the first one by parts?
Yes, there were a couple of typos in HoI's reply which you have spotted and fixed.

See integrate x &#40;Sin&#91;x&#93;&#41;&#94;2 Cos&#91;x&#93; - Wolfram|Alpha and click on Show steps.