Is it true $\int_0^k \frac{\partial f(x,y)} {\partial x}\,dx = f(k,y) + \textrm{a function of }y$?

Question

I saw it written somewhere that

$$\int_0^k \frac{\partial f(x,y)} {\partial x}\,dx =f(k,y) + \textrm{a function of }y$$

This seems feasible, but I haven't seen the integral of a partial derivative before. Can someone prove the above statement?

Answer 1

Ignore the $y$ variable for the time being -- it turns out the fact that $f$ is multivariate is not the important aspect here. The important thing to realize here, is that the initial limits on the integral are from $x=0$ to $x=k$. When one cancels the $\dfrac{dx}{dx}$, the integral is now with respect to $df(x)$ and the limits must be changed to be from $f(x)=f(0)$ to $f(x)=f(k)$.

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In other words,

$$\begin{align} \int_0^k \dfrac{\partial f(x,y)} {\partial x}dx &\equiv \int_{x=0}^{x=k} \dfrac{\partial f(x,y)} {\partial x}dx\\ &= \int_{f(x,y)=f(0,y)}^{f(x,y)=f(k,y)} df(x,y)\\ &\equiv \int_{\xi=f(0,y)}^{\xi=f(k,y)} d\xi\\ &= \xi|_{f(0,y)}^{f(k,y)}\\ &= f(k,y)-f(0,y) \end{align}$$ by the fundamental theorem of calculus.

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Another way to evaluate the integral is: $$ \int_0^k \dfrac{\partial f(x,y)} {\partial x}dx \equiv \int_{x=0}^{x=k} \dfrac{\partial f(x,y)} {\partial x}dx = f(x,y)|_{x=0}^{x=k} = f(k,y)-f(0,y) $$

This second way is clearly the "easier" derivation, but it is important to understand the first way so that one understands what is going on behind the scenes (and also to see that $\int \dfrac{\partial f(x,y)} {\partial x}dx = f(x,y)$ where the limits only affect the $x$ variable)