Question

impulse function:

Integral:

Answer 1

\[\int\limits_{-4}^{4} 3(t-2)^{2}\delta'(t-3/2) dt\]

Answer 2

Ia this the derivative of the dirac delta function inside the integral?

Answer 3

its the first derivative of the delta function inside the integral

Answer 4

3

Answer 5

I've got \(3\) too:
\[\int\limits_{-4}^{4}3(t-2)^2δ'(t-{3 \over 2})dt=-\int\limits_{-4}^{4}6(t-2)δ(t-\frac{3}{2})=-6(\frac{3}{2}-2)=-6(\frac{-1}{2})=3\]

Answer 6

but that isnt right.. its not a "normal" integral... you have to solve this with partial integration or somthing like that

Answer 7

http://www.wolframalpha.com/input/?i=Integrate [DiracDelta%27[t-3%2F2]*3%28t-2%29^2%2C+{t%2C+-4%2C+4}]

Answer 8

http://alturl.com/7ibic

Answer 9

The formula you need to use here is:
\[\int\limits_{-\infty}^{\infty}f(x)δ'(x)dx=\int\limits_{-\infty}^{\infty}f'(x)δ(x)dx.\]

Answer 10

I guess you know that \(\int_{-\infty}^{\infty}f(x)δ(x-a)=f(a).\)

Answer 11

yeah i know that one :)

Answer 12

Why are you saying that \(5e^{-24}\) is the answer?

Answer 13

my book says this:

\[\int\limits_{-4}^{4}[(3/4)\delta'(t-(3/2) + 9\delta(t-(3/2)]dt\]=

\[(3/4)\delta(t-(3/2) + 9\]

Answer 14

the 5e^-24 is my bad.. thats another question ;)

Answer 15

i dont understand the \[9\delta(t-(3/2))\]

Answer 16

I'm confused! What exactly don't you understand?

Answer 17

oke.. sorry i confused you, but that i typed "my books says". thats the solution of my question.

they use partial integration. but the second term. 9 delta(t-(3/2)) there does this term comes from

Answer 18

Sorry, I don't know what your book is doing. You probably mean integration by parts, but even that would look different from what you wrote.