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

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?

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

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• – Baklava Gain Jun 11 at 11:27

If $$x$$ is fixed, then $$f(x,s)$$ is an anti- derivative of $$\frac{\partial}{\partial s} f(x,s)$$. Hence
$$\int_{0}^{t }\left(\frac{\partial}{\partial s} f(x,s)\right)ds =f(x,t)-f(x,0).$$