How can we simplify the integral of a step function with a derivative?

[courtrigrad]

Suppose U(x) is the step function and \delta(x) is its derivative. Find \int^{6}_{-2}(x^{2}-8)\delta(x)\dx. I know \delta(x) = 0 for all x except x = 0. So at x = 0 v(x) = - 8. After this step, how do we get \int_{-2}^{2}(x^{2}-8)\delta(x)\dx +\int^{6}_{2}(x^{2}-8)\delta(x)\dx = -8 + 0. How do you get the limits of integration?

Thanks

 

[benorin]

If by step function you mean

U(x)=\left\{\begin{array}{cc}0,&\mbox{ if }x\leq 0\\1, & \mbox{ if } x\geq 0\end{array}\right.​

(a.k.a. the Heaviside step function) then \frac{d}{dx}U(x) = \delta (x) is a Dirac delta function which has the so-called snifting property, that is

\int_{-\infty}^{\infty}f(x)\delta (x) \, dx = f(0)​

so the limits of integration don't matter (so long as they contain zero) and the value of the integral is -8.

 

[FrogPad]

Suppose U(x) is the step function and \delta(x) is its derivative. Find \int^{6}_{-2}(x^{2}-8)\delta(x)\dx.
Thanks

Remember that you can split up the integral. So,

\int^{6}_{-2}(x^{2}-8)\delta(x)\dx = \int_{-2}^{0}(x^{2}-8)\delta(x)\dx = \int_{0}^{6}(x^{2}-8)\delta(x)\dx

\int^{6}_{-2}(x^{2}-8)\delta(x)\dx = \int_{-2}^{0}(x^{2}-8)\delta(x)\dx = \int_{0}^{1}(x^{2}-8)\delta(x)\dx =\int_{1}^{6}(x^{2}-8)\delta(x)\dx

or however you want (within the bounds of the original limits, etc...). So the integral equals 0 for some limits of integration, it will hit 0 and the sifting property kicks in, and then the integral equals 0 again.