Question

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

Answer 1

14

Answer 2

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

Answer 3

What type of couse is this for? Just curious

Answer 4

Darn, I was going about it all wrong lol.. And it's for a linear algebra/differential equations class

Answer 5

Seems interesting

Answer 6

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

Answer 7

oops, it's supposed to be g sub h of t

Answer 8

What function is g sub h?

Answer 9

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

Answer 10

The limit part. I guess the dirac function is the derivative of the heavyside

Answer 11

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}]

Answer 12

Sorry about the double posting. Using Firefox and it seems to have a hair trigger.