Problem with integrating heaviside function.

[perpich08]

Homework Statement

I need to find the integral of the following equation..

integral from 0 to 8 of {(2H[x-0]+2H[x-4])*(x/8)}dx

The Attempt at a Solution

Im am not sure what the integral of the heaviside function is?
is it 1 or 0?

Any help would be appreciated!
Thanks

 

[LCKurtz]

Write your function as a piecewise defined function and you will see how to integrate it.

f(x) = ? if 0 < x < 4
f(x) = ? if 4 < x < 8

Remember H(x) = 0 if x < 0 and H(x) = 1 if x > 0.

Also drawing the graph might help.

 

[HallsofIvy]

The integral of the Heaviside function, H(x), is just xH(x)+ C.

 

[Dickfore]

Im am not sure what the integral of the heaviside function is?

Use integration by parts and the fact that h&#039;(x - x_{0}) = \delta(x - x_{0}):

<br /> \int_{a}^{b}{h(x - x_{0}) \, dx} = \left. x \, h(x - x_{0}) \right|^{b}_{a} - \int_{a}^{b}{x \, \delta(x - x_{0}) \, dx}<br />

Next, the integral with the Dirac delta-function can be evaluated using its property:

<br /> \int_{a}^{b}{x \, \delta(x - x_{0}) \, dx} = x_{0} \, h(x_{0} - a) \, h(b - x_{0})<br />

where the product of the two Heaviside functions ensures that x_{0} is inside the segment x_{0} \in \left[a, b \right]. The integrated-out part is:

<br /> b \, h(b - x_{0}) - a \, h(a - x_{0})<br />

Combining everything together gives:

<br /> \int_{a}^{b}{h(x - x_{0}) \, dx} = b \, h(b - x_{0}) - a \, h(a - x_{0}) - x_{0} \, h(x_{0} - a) \, h(b - x_{0})<br />

 

[Dickfore]

As a check, let us differentiate it w.r.t. b:

<br /> \frac{\partial}{\partial b} \left( b \, h(b - x_{0}) \right) = 1 \cdot h(b - x_{0}) + b \cdot \delta(b - x_{0}) = h(b - x_{0}) + x_{0} \, \delta(b - x_{0})<br />

<br /> \frac{\partial}{\partial b} \left( -a \, h(a - x_{0}) \right) = 0<br />

<br /> \frac{\partial}{\partial b} \left( -x_{0} \, h(b - x_{0}) h(x_{0} - a) \right) = -x_{0} \, \delta(b - x_{0}) \, h(x_{0} - a) = -x_{0} \, \delta(b - x_{0}) \, h(b - a) = -x_{0} \delta(b - x_{0})<br />

where again we used the derivative of the Heaviside step function and a consequence of the properties of the Dirac delta-function (x \, \delta(x - x_{0}) = x_{0} \, \delta(x - x_{0})) as well as the assumption that b &gt; a. Combining everything together and substituting b \rightarrow x. we see that we get the integrand. Also, if b = a, the first two terms cancel and the last cancels because h(a - x_{0}) h(x_{0} - a) is always zero. So, according to the fundamental theorem of calculus, we should get the correct result.

 

[Dickfore]

Of course, you can always write:

<br /> \int_{0}^{8}{\left( 2 h(x) + 2 h(x - 4) \right) \frac{x}{8} \, dx} = \frac{1}{4} \, \left[\int_{0}^{8}{x \, h(x) \, dx} + \int_{0}^{8}{x \, h(x - 4) \, dx}\right]<br />

<br /> \frac{1}{4} \, \left[ \int_{0}^{8}{x \, dx} + \int_{4}^{8}{x \, dx}\right]<br />

and do the remaining elementary integrals :)

 

[perpich08]

Thanks guys! I think I can figure it out now

 

[HallsofIvy]

I would consider using integration by parts for integrating the Heaviside function "overkill".

If a<0, then H(x)= 0 for all x\le 0 so \int_{-\infty}^a H(x)dx= 0. For a\ge 0, \int_{-\infty}^a H(x)dx= \int_0^a H(x)dx= \int_0^a dx= a.

For the "indefinite integral" \int H(x)dx is 0+ C for x< 0, x+ C for x\ge 0.