We have an answer to "f(x,y)=∫yxg(t)dt g(t) continuous for all t. Evaluate the..."

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eozoischgc

Answered question

2021-12-07

$f (x, y) = \int_{y}^{x} g (t) d t$
g(t) continuous for all t. Evaluate the derivatives of f.

Answer & Explanation

ramirezhereva

Beginner2021-12-08Added 28 answers

Let a be a real. Thus,
$f (x, y) = \int_{a}^{x} g (t) d t - \int_{a}^{y} g (t) d t = G (x) - G (y)$
$G (X) = \int_{a}^{X} g (t) d t$
$G^{'} (X) = g (X)$
Since g is continuous at R,
$f_{x} (x, y) = G^{'} (x) = g (x)$
$f_{y} (x, y) = G^{'} (y) = - g (y)$

Kindlein6h

Beginner2021-12-09Added 27 answers

$\frac{d}{d x} \int_{α (x)}^{β (x)} f (x, t) d t = f (x, β (x)) - f (x, α (x)) + \int_{α (x)}^{β (x)} \frac{\partial f (x, t)}{\partial x} d t$
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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