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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
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

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jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1355956366325:dw \[\int\limits_{2}^{1} xf(x)dx\]

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.1basically the pieces \[x, x \le 0\] \[x+1, x > 0\]

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.1i 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...

beginnersmind
 3 years ago
Best ResponseYou've already chosen the best response.1Limits of integration should be [2,0] and [0,1] respectively.

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.1ohhh that's what i thought but cuz it's not greater than or equal to, i didn't put the zero...

beginnersmind
 3 years ago
Best ResponseYou've already chosen the best response.1Also, for the first part you seem to have taken the integral from 0 to 2. It should be from 2 to 0.

jennychan12
 3 years ago
Best ResponseYou've already chosen the best response.1ok yeah thanks. i got 7/2 as the answer

beginnersmind
 3 years ago
Best ResponseYou've already chosen the best response.1That's what I got too.

agent0smith
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits\limits_{1}^{1} x(x+1)dx \]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..
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