- #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...
Definite integration is a mathematical concept used to find the exact area under a curve between two specific points on a graph. It involves finding the antiderivative of a function and evaluating it at the upper and lower limits of integration.
Definite integration involves finding the exact area under a curve between two specific points, while indefinite integration involves finding the general antiderivative of a function. In other words, definite integration gives a specific numerical value, while indefinite integration gives a function.
Definite integration is used in various fields of science, engineering, and economics to find the total value, area, volume, or other quantities that can be represented by a function. It is also used to solve problems involving rates of change and accumulation.
The basic steps for evaluating a definite integral are: 1) finding the antiderivative of the function, 2) substituting the upper and lower limits of integration into the antiderivative, 3) subtracting the two values to find the difference, and 4) simplifying the result if possible.
Yes, definite integration can be used for functions with multiple variables, but it requires using multiple integrals. The process is similar to evaluating a single integral, but with additional steps for each variable. This is commonly used in physics and engineering to find the volume, surface area, and other quantities in three-dimensional space.