Integral depending on a parameter

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SUMMARY

The discussion focuses on finding the derivative I'(y) of the integral I(y) = ∫₀ʸ f(x,y)dx. The solution involves defining an auxiliary function J(t,u) = ∫₀ᵗ F(x,u)dx and applying the Fundamental Theorem of Calculus (FTC) along with the Multivariable Chain Rule. The use of Leibniz's rule is emphasized as a more general approach for differentiating under the integral sign, which incorporates the partial derivatives of F with respect to y.

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Homework Statement



Let I(y) = [tex]\int_0^y f(x,y)dx[/tex].
Find I'(y).

Homework Equations





The Attempt at a Solution



I have the solution which involves making an auxiliary function [tex]J(t,u) = \int_0^tF(x,u)dx[/tex] and expressing I'(y) in terms of the partial derivatives of J(t,u). It uses the FTC and the Multivariable Chain Rule. However, it is extremely confusing and if someone could break it down for me that would be great. :)
 
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You could use the chain rule. Or you could use Leibniz's rule which is slightly more general than that:

[tex]\frac{d}{dy}\int_{\alpha(y)}^{\beta(y)} F(x,y)dx= \int_{\alpha(y)}^{\beta(y)}\frac{\partial F}{\partial y}dx+ \frac{d\alpha}{dy}F(\alpha(y),y)-\frac{d\beta}{dy}F(\beta(y),y)[/tex]
 

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