i am doing a problem by parts, and my dv= e^-2t dt i was just wondering how one knows that v, then, =-1/2 e^-2t
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Originally Posted by buttonbear i am doing a problem by parts, and my dv= e^-2t dt i was just wondering how one knows that v, then, =-1/2 e^-2t v=-1/2 e^-2t because the derivative of v is dv. v=-1/2 e^-2t then $\displaystyle dv=e^{-2t} \cdot -2 \cdot \frac{-1}{2}=dv= e^{-2t}$ That's why.
i know that the derivative of v is dv i'm sorry if this is a stupid question but if i'm starting with dv, i just don't see how you integrate dv to get v..
Originally Posted by buttonbear i am doing a problem by parts, and my dv= e^-2t dt i was just wondering how one knows that v, then, =-1/2 e^-2t b/c $\displaystyle \int e^{-2t}dt$ $\displaystyle =e^{-2t} \cdot \frac{1}{-2}+C$ and double check you have the right integral by differentiating it. there are many steps involving $\displaystyle \int e^{f(x)}dx$
Originally Posted by buttonbear i know that the derivative of v is dv i'm sorry if this is a stupid question but if i'm starting with dv, i just don't see how you integrate dv to get v.. $\displaystyle \int e^{kx}~dx = \frac 1k e^{kx} + C$, for $\displaystyle k \ne 0$ a constant. we obtain this by doing a substitution of $\displaystyle u = kx$
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