eozoischgc

2021-12-07

$f\left(x,y\right)={\int }_{y}^{x}g\left(t\right)dt$
g(t) continuous for all t. Evaluate the derivatives of f.

ramirezhereva

Let a be a real. Thus,
$f\left(x,y\right)={\int }_{a}^{x}g\left(t\right)dt-{\int }_{a}^{y}g\left(t\right)dt=G\left(x\right)-G\left(y\right)$
$G\left(X\right)={\int }_{a}^{X}g\left(t\right)dt$
${G}^{\prime }\left(X\right)=g\left(X\right)$
Since g is continuous at R,
${f}_{x}\left(x,y\right)={G}^{\prime }\left(x\right)=g\left(x\right)$
${f}_{y}\left(x,y\right)={G}^{\prime }\left(y\right)=-g\left(y\right)$

Kindlein6h

$\frac{d}{dx}{\int }_{\alpha \left(x\right)}^{\beta \left(x\right)}f\left(x,t\right)dt=f\left(x,\beta \left(x\right)\right)-f\left(x,\alpha \left(x\right)\right)+{\int }_{\alpha \left(x\right)}^{\beta \left(x\right)}\frac{\partial f\left(x,t\right)}{\partial x}dt$
f(x,t) is a function of t only, the upper bound on the integral is just x and the lower bound just y. So the derivative of the integral with respect to x is g(x) and the derivative with respect to y is −g(y).

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