Determining the velocity is terms of displacement given acceleration. (terminology)

Question

So I need to determine the velocity in terms of displacement. It was given the acceleration of the particle is $a(x) = 49x - 14$

I know that $a = \frac{d}{dx}(\frac{1}{2}v^2)$; which is $49x-14\:=\:\frac{d}{dx}\left(\frac{1}{2}v^2\right)$

This is where I got bamboozled, my textbook went straight to $\frac{1}{2}v^2=\int \:49x-14\:dx$. How?

I get that you need to times $dx$ by both sides. But Where did the $"d"$ go? So when you integrate both sides, the $"d"$ and the integral sign cancel out?

Like so: $\int \:\frac{1}{2}v^2\:\:\:d=\int \:49x-14\:dx$

and $"d"$ cancels out?

$\:\frac{1}{2}v^2\:=\int \:49x-14\:dx$

Answer 1

In general we have that by definition:

$$\frac {dF(x)}{dx}=f(x) \iff F(x)= \int_0^x f(u) du$$

with $F(x)=\frac12 v^2$ and $f(x)=49x-14$ in your case.

We don't cancel out "d" which is just used as a symbol to indicate the derivative or the integrating variable.

Answer 2

Note that

$$\frac{d}{dx}\left(\frac{1}{2}v^2\right) =49x-14 $$

leads to $$d\left(\frac{1}{2}v^2\right) =(49x-14)dx $$

Then, integrate both sides

$$\int d\left(\frac{1}{2}v^2\right)= \frac{1}{2}v^2 =\int (49x-14)dx $$