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jennychan12

  • 2 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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  1. jennychan12
    • 2 years ago
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    |dw:1355956366325:dw| \[\int\limits_{-2}^{1} xf(x)dx\]

  2. jennychan12
    • 2 years ago
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    basically the pieces \[x, x \le 0\] \[x+1, x > 0\]

  3. jennychan12
    • 2 years ago
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    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
    • 2 years ago
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    Limits of integration should be [-2,0] and [0,1] respectively.

  5. jennychan12
    • 2 years ago
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    ohhh that's what i thought but cuz it's not greater than or equal to, i didn't put the zero...

  6. beginnersmind
    • 2 years ago
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    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
    • 2 years ago
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    oops my bad.

  8. jennychan12
    • 2 years ago
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    ok yeah thanks. i got 7/2 as the answer

  9. beginnersmind
    • 2 years ago
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    That's what I got too.

  10. jennychan12
    • 2 years ago
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    k thanks.

  11. agent0smith
    • 2 years ago
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    \[\int\limits\limits_{1}^{1} x(x+1)dx \]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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