Change of Variable issue with Integration

[dimensionless]

I have the following equation:

<br /> I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right], dx<br />

I have set

<br /> y= \frac{x}{a}-\frac{1}{2}<br />

and

<br /> dy = dx/a<br />

When I substitute the two latter equations into the first equation I should get this:

<br /> I = {a} \int_{-1}^{0} (2y+1) sin^{2}\left[ \pi y\right], dy<br />

For some reason I get this instead:

<br /> I = \frac{a}{2} \int_{-1}^{0} (y+\frac{1}{2}) sin^{2}\left[ \pi y\right], dy<br />

I'm off by a factor of four. What am I doing wrong?

 

[Hammie]

It looks like you substitued y = x/a + 1/2 in the integral, and used x/a - 1/2 to determine the new limits of integration.

<br /> I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}+\frac{1}{2})\right], dx<br />

Or should have this been

<br /> I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right], dx<br />

 

[dimensionless]

The given equation should have been

<br /> I = \frac{a}{2} \int_{-a/2}^{a/2} x sin^{2}\left[ \pi (\frac{x}{a}-\frac{1}{2})\right] dx<br />

I have corrected this in the initial post.

Unfortunately I'm still stuck same answer (and this answer does not match the one in my book).

 

[marlon]

If y= \frac{x}{a}-\frac{1}{2} then x = a(y + \frac{1}{2})

and

dx = ady

So we get for the integral

I = \frac{a}{2} \int_{-1}^{0} a(y + \frac{1}{2}) sin^{2}\left[ \pi y \right] ady

marlon

 

[dimensionless]

Well, this book does have some errors in it.