# How integrate with respect to a variable affect the whole expression

I have spent some time to solve this equation and I don't know how to solve it: $$\int_0^t e^{-rs}dk_s ,$$ where r is constant. I was thinking the result would be $$e^{-rs}k_s$$. But should the exponential term affect too?

Ultimately, I want to take the derivative of the above expression with respect to t. Seems like the answer is $$e^{-rt}dk_t$$. Could anyone please explain how to arrive at the answer? Thanks!

• I suppose you know what $dk_s$ means. If not, sit down and recall the definition and how to obtain $ds$ from $dk_s$. If you know this, you can easily solve the integral. – user526015 Feb 19 at 10:34
• Hi, thanks for the reply. Unfortunately, there is no definition about $dk_s$, so I can't find a way to restate $dk_s$ in terms of $ds$. – U_ng Feb 19 at 10:56

I will write $$k(s)$$ instead of $$k_s$$. Using "physicists notation", $$\frac{dk(s)}{ds}=k'(s) \Longleftrightarrow dk(s)=k'(s)ds.$$ So you have to calculate $$\int_0^t e^{-rs}dk(s)=\int_0^te^{-rs}k'(s)ds.$$ Using the Fundamental Theorem of Calculus, this is $$F(t)-F(0)$$ where $$F$$ is a primitive function for the integrand. Thus, differentiating w.r.t. $$t$$ should give back the integrand, i.e. $$e^{-rt}k'(t).$$