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Why is the following true?

$$\int_{0}^{t }\left(\frac{\partial}{\partial s} f(x,s)\right)ds =f(x,t)$$

Can you give any hints?

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closed as off-topic by YuiTo Cheng, Daryl, Jendrik Stelzner, José Carlos Santos, Aweygan Jun 11 at 14:55

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If $x $ is fixed, then $f(x,s)$ is an anti- derivative of $\frac{\partial}{\partial s} f(x,s)$. Hence

$\int_{0}^{t }\left(\frac{\partial}{\partial s} f(x,s)\right)ds =f(x,t)-f(x,0).$

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