# Delta function integral

let $$f(y)=\int_0^2 \delta(y-x(2-x))dx$$. Find f(y) and plot it from -2 to 2.

I know how to calculate $$\delta (g(x))$$ but i am not sure how to treat it with the y. I thought possibly to solve the quadratic in the delta function to find what x will equal for the roots in terms of y and got 1+(1-y)^1/2 and 1-(1-y)^1/2. i am not sure though how to find f(y) with the integral. any suggestions?

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arildno
Homework Helper
Gold Member
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Treat the cases separately:
1. The singularity lies in the open interval between the end points
2. The singularity lies outside the interval
3. The singularity lies at an end point

HallsofIvy
Homework Helper
The singularity arildno refers to is where y- x(2-x)= 0. In other words when y= 2x- x2. Find the value of that integral, f(y), for y in the positions arildno listed and that's your function.

so what youre saying is to form the integral $$f(y)=\int 2x-x^2 dx$$?
if its outside the interval do i let the limits be (-infinity to 0) and (2 to infinity)
for inside 0 to 2 but at the endpoints what are the limits?

arildno
Homework Helper
Gold Member
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No, why do you think that?

HallsofIvy
Homework Helper
What is $\int_a^b \delta(x)dx$ for any a, b?

D H
Staff Emeritus
What is $\int_a^b \delta(x)dx$ for any a, b?
Even more appropriately,

What is $\int_a^b \delta(x)f(x)dx$ for any a, b?

One way to solve the problem at hand is to convert the original problem,
$$f(y)=\int_0^2 \delta(y-x(2-x))dx$$
to the form
$$f(y)=\int_{u_0}^{u_1} \delta(u)f(u)du$$

Note that the integrand may have two singularities between the integration limits, depending on the value of y.