- #36
Raghav Gupta
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- 76
I think I am using a mobile device and you desktop.certainly said:It's visible here, @Ray Vickson is it not visible to you too...
I think I am using a mobile device and you desktop.certainly said:It's visible here, @Ray Vickson is it not visible to you too...
certainly said:It's visible here, @Ray Vickson is it not visible to you too...
Raghav Gupta said:I think I am using a mobile device and you desktop.
Raghav Gupta said:Isn't Leibniz rule
$$ \frac{d}{dx} \int_{g(x)}^{h(x)} f(t) = f(h(x))h'(x) - f(g(x)) g'(x) $$ ?
So it should be
$$ \frac{d}{dx} \int_0^x e^{-t} f(x-t) = e^{-x}f(0) ? $$ as second term would be zero because differentiation of a constant which is the lower limit is zero
Problems arising for me.certainly said:No the leibniz integral rule is:-
$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}f(x,t)\ dt=\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}\ dt+f(b(x),x)\frac{\partial b}{\partial x}-f(a(x),x)\frac{\partial a}{\partial x}$$
i'm going to call the function to be integrated ##g(x,t)## (i.e ##f## in the equation you quoted) because it might cause some confusion.Raghav Gupta said:Problems arising for me.
Getting first term in R.H.S
∫x0e−tf′(x−t)
\int_0^x e^{-t} f'(x-t) second term 0 and third term also 0 as δa/δx is 0.
No.. it wasn't mentioned in the question.certainly said:It is incorrect to assume that ##f(0)=0##...
And if it was mentioned in the question, you should have told us so...
But definitely f(0) = 0mooncrater said:Homework Statement
The question says:
f (x)=x2+7x +∫0x(e-tf (x-t)dt.
Find f (x).
For you .certainly said:errrrr... yes, sorry for that, looks like I'm finally getting sleepy...