G(x) is the antiderivative of F(x) with G(1) = 1; Find a formula for G(x)

[IntegrateMe]

G(x) is the antiderivative of F(x) with G(1) = 1; Find a formula for G(x)...

Find a formula for G(x) in terms of F(x).

I would have just written \int_0^x{F(x)dx} but I know the value G(1) = 1 has to be included somewhere. Can someone help me figure this one out?

 

[Dick]

How about if you just add a constant?

 

[IntegrateMe]

\int_0^x{F(x)dx} + C??

 

[Dick]

\int_0^x{F(x)dx} + C??

Sure. If that is your G(x) then what value of C will make G(1)=1? Put x=1 and solve for C.

 

[IntegrateMe]

Well, wouldn't I need to know what the value of \int_0^1{F(x)dx} is to determine that?

 

[Dick]

Well, wouldn't I need to know what the value of \int_0^1{F(x)dx} is to determine that?

No. You can't compute it. \int_0^1{F(x)dx} should appear as part of your answer.

 

[IntegrateMe]

G(x) = 1 + \int_1^xF(s)ds ?

 

[Dick]

G(x) = 1 + \int_1^xF(s)ds ?

Good job. Why the '?'

 

[IntegrateMe]

I wasn't sure of the limits of integration. They were kind of unjustified.

 

[Dick]

I wasn't sure of the limits of integration. They were kind of unjustified.

If you'd followed the line I was suggesting you should have found C=1-\int_0^1{F(s)ds}

Can you justify the bounds using that?

 

[IntegrateMe]

The F(x) was making it confusing.

 

[Dick]

The F(x) was making it confusing.

Maybe. If one of the limits of integration is x you shouldn't write the integral dx. Since the integration variable is a dummy you should use something besides x, like s. Makes life less confusing.

 

[IntegrateMe]

Can you explain why C would allow you to justify the bounds.

 

[Dick]

The integral from 0 to x of F minus the integral from 0 to 1 of F equals the integral from 1 to x of F.