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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
 one year ago
 one year 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
 one year ago
 one year ago

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jennychan12Best ResponseYou've already chosen the best response.1
dw:1355956366325:dw \[\int\limits_{2}^{1} xf(x)dx\]
 one year ago

jennychan12Best ResponseYou've already chosen the best response.1
basically the pieces \[x, x \le 0\] \[x+1, x > 0\]
 one year ago

jennychan12Best ResponseYou've already chosen the best response.1
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...
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
Limits of integration should be [2,0] and [0,1] respectively.
 one year ago

jennychan12Best ResponseYou've already chosen the best response.1
ohhh that's what i thought but cuz it's not greater than or equal to, i didn't put the zero...
 one year ago

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

jennychan12Best ResponseYou've already chosen the best response.1
ok yeah thanks. i got 7/2 as the answer
 one year ago

beginnersmindBest ResponseYou've already chosen the best response.1
That's what I got too.
 one year ago

agent0smithBest 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..
 one year ago
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