Definite Integral of a derivative

Question

I am doing a differential equation for my physics class, and I have gotten to $$\frac{d}{dt}\left[e^{\text{kt}} v_y(t)\right]=-g e^{\text{kt}}$$ and just want to perform a definite integral. $$\int_{v_y(0)}^{v_{y}(t)} d\left[e^{\text{kt}} v_y(t)\right]=\int_{0}^{t} -g e^{\text{kt}} \, dt$$

When the derivative is undone by the integral, which variable do I evaluate for the bounds on the left integral?

Answer 1

Call the variable $u(t,v_y) = e^{kt}v_y$, so the LHS of your last equation becomes

$$ \int_{u(0,v_{y_0})}^{u(t,v_{y})}{\rm d}u = u(t,v_{y}) - u(0,v_{y_0}) $$

That is, the variable is $u$

Answer 2

It's simply

$$\int_{v_{y_0}}^{v_{y}} d\left[e^{\text{kt}} v_y(t)\right]=(e^{\text{kt}} v_y(t))\big |_{t=v_{y_0}}^{t=v_{y}}$$ Like in $$\int_a^b dx^x=(x^x)\big |_a^b=b^b-a^a$$

Normally you should have $$\frac{d}{dt}\left[e^{\text{kt}} v_y(t)\right]=-g e^{\text{kt}}$$ $$\int_0^t d(e^{\text{kt}} v_y(t))=-\int_0^tg e^{\text{kt}}dt$$

Answer 3

Your bounds on the left are wrong.

Assuming $k \ne 0$ and $g$ are constant regarding time, we have $$ \frac{d}{dt} \left(e^{kt} v_y(t)\right) = -g\,e^{kt} \iff \\ \left[ e^{k\tau} v_y(\tau) \right]_{\tau=t_0}^{\tau=t} = -g \int\limits_{t_0}^t e^{k\tau} d\tau \iff \\ e^{kt} v_y(t) - e^{kt_0} v_y(t_0) = -\frac{g}{k}\left(e^{kt}-e^{kt_0}\right) $$ where we integrate both sides over time (using a new integration variable $\tau$ to avoid a name clash with the upper bound) from $t_0$ to $t$.