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Is the solution of a generic integral of the form $$ \int_0^t \frac{\partial f(x(y),y)}{\partial y} dy $$ where $x(y)$ is unknown, simply equal to $$f(x(t),t) - f(x(0),0)?$$ And if it is not always the case, when does it hold? I previously found a related question here.

Mr Frog
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2 Answers2

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The formula $$ \int_0^t\partial_yf(x(y),y)\,dy=f(x(t),t)-f(x(0),0) $$ holds for all $t$ if and only if

  • $x$ does not depend on $y\,,$ or, $f(x,y)$ does not depend on $x\,.$

Proof. Differentiating the formula with respect to $t$ gives $$ \partial_yf(x(t),t)=x'(t)\partial_xf(x(t),t)+\partial_yf(x(t),t)\,. $$ This can be simplified to $$ 0=x'(t)\partial_xf(x(t),t)\,. $$ So either $x'(t)=0$ or $\partial_xf(x(t),t)=0\,.$

The reverse implication is shown in the linked question for $x$ being constant. When $f$ does not depend on $x$ it is the well known fundamental theorem of calculus.

Kurt G.
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  • In case x did not depend on y, the formula would hold up to a constant that does not depend on y but might depend on x, right? – Mr Frog Jan 30 '22 at 20:25
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    In the link they consider an indefinite integral wich brings up that constant. We here consider a definite integral from zero to $t,.$ No constant. – Kurt G. Jan 30 '22 at 20:32
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By the fundamental theorem of calculus, $$\int_0^t\frac{\mathrm{d}f(x(y),y)}{\mathrm{d}y}\,\mathrm{d}y=f(x(t),t)-f(x(0),0).$$ To see why, just let $g(y):=f(x(y),y),$ and proceed from there. By the chain rule, $$\frac{\mathrm{d}f(x(y),y)}{\mathrm{d}y}=\frac{\partial{f(x(y),y)}}{\partial{y}}+\frac{\partial{f(x(y),y)}}{\partial{x}}x'(y).$$ Therefore, $$\int_0^t\frac{\partial{f(x(y),y)}}{\partial{y}}\,\mathrm{d}y=f(x(t),t)-f(x(0),0)-\int_0^t\frac{\partial{f(x(y),y)}}{\partial{x}}x'(y)\,\mathrm{d}y.$$ This is the formula you need. Whether this is helpful or not is a separate subject of discussion, of course.

Angel
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