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anonymous

  • 5 years ago

Evaluate the integral from -infinity to infinity of [(t^2+5t)*dirac(t-2)dt]

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  1. anonymous
    • 5 years ago
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    14

  2. anonymous
    • 5 years ago
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    Wiki says that the intergral from -infinity to infinity of f(t)dirac(t-T)dt is equal to f(T). So i guess if you distribute and split up the integral it becomes (2)^2+5(2)=14

  3. anonymous
    • 5 years ago
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    What type of couse is this for? Just curious

  4. anonymous
    • 5 years ago
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    Darn, I was going about it all wrong lol.. And it's for a linear algebra/differential equations class

  5. anonymous
    • 5 years ago
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    Seems interesting

  6. anonymous
    • 5 years ago
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    In my textbook it's talking about how dirac(t) = lim h -> 0 [g(h(t)] and then you plug it into the general equation, but I don't really get the whole limit part

  7. anonymous
    • 5 years ago
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    oops, it's supposed to be g sub h of t

  8. anonymous
    • 5 years ago
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    What function is g sub h?

  9. anonymous
    • 5 years ago
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    oh whoops, I forgot to add that part, it's a heavyside function. g sub h = (1/h)[H(t)-H(t-h)]

  10. anonymous
    • 5 years ago
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    The limit part. I guess the dirac function is the derivative of the heavyside

  11. anonymous
    • 5 years ago
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    If "dirac" is the DiracDelta function then, \[\int\limits _{-\infty }^{\infty }\left(5 t+t^2\right) \delta (-2+t)dt=14 \] Browse over to WolframAlpha.com and paste in the following: Integrate[(t^2 + 5 t) DiracDelta[-2 + t], {t, -Infinity, Infinity}]

  12. anonymous
    • 5 years ago
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    Sorry about the double posting. Using Firefox and it seems to have a hair trigger.

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