## jennychan12 3 years ago Let f(x) be defined by f(x) = x if x</= 0 and x+1 if x> 0. the value of the integral from -2 to 1 of xf(x)dx = a.3/2 b.5/2 c.3 d. 7/2 e. 11/2

1. jennychan12

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

2. jennychan12

basically the pieces $x, x \le 0$ $x+1, x > 0$

3. jennychan12

i think you supposed to break it up? $\int\limits_{1}^{1} x(x+1)dx$ = 0 $\int\limits_{0}^{-2} x^2dx$ = -8/3 i'm doing something wrong...

4. beginnersmind

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

5. jennychan12

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

6. beginnersmind

Also, for the first part you seem to have taken the integral from 0 to -2. It should be from -2 to 0.

7. jennychan12

8. jennychan12

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

9. beginnersmind

That's what I got too.

10. jennychan12

k thanks.

11. agent0smith

$\int\limits\limits_{1}^{1} x(x+1)dx$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..