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|dw:1355956366325:dw|
\[\int\limits_{-2}^{1} xf(x)dx\]

basically the pieces
\[x, x \le 0\]
\[x+1, x > 0\]

Limits of integration should be [-2,0] and [0,1] respectively.

ohhh that's what i thought but cuz it's not greater than or equal to, i didn't put the zero...

oops my bad.

ok yeah thanks. i got 7/2 as the answer

That's what I got too.

k thanks.

\[\int\limits\limits_{1}^{1} x(x+1)dx \]Looks like you got this already, but as for this part with the limits of integration that you were confused on - you can still pick the first limit on the integral as 0, even though the function is defined as x>0.
The picture might help. Even though x>0, the area under the curve is still effectively from x=0 to x=1. If you only integrate from 1 to 1, the area under that is zero. You could pick a value of x just slightly greater than x=0, like 0.0000000000001, and do the integration, and you'll get essentially the same amount for the area from x=0.0000000000001 to x=1 as you do for x=0 to x=1.
Sorry if that sounded confusing..