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I'm reading this book: Modern Elementary Differential Equations

I understand that the second derivative of x (position) with respect to t is the downward effect gravity (acceleration) has on the ball. I understand the initial conditions of position being 0 and velocity being whatever it is initially (5.4). I understand the integration result in 5.5 But I don't understand how putting together 5.4 and 5.5 leads to 5.6.

Why does

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Become

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Here is the text from the book for reference:

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    Am I overthinking this? Are they really just saying the integration *IS* velocity? – 2017-01-01
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    That is to say, that if you remove the initial velocity from the equation, all you're left with is the effect of gravity. – 2017-01-01
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    To me, it seems like if you integrate first part of 5.5 then it would result in dx/dt - v by fundamental theorem of calculus. F[t] - F[0] with F(0) = x'(0) = v. F[t] is simply dx/dt. – 2017-01-01

2 Answers 2

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If you integrate $\frac{d^2x}{dt^2}$, by fundamental theorem of calculus it would be

F[t] - F[0] = $x'(t)$-$x'(0)$ = $\frac{dx}{dt}$ - v

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They are using the (slightly confusing) shorthand of simply writing $\frac{dx}{dt}$ when what they really mean is $\frac{dx}{dt}\bigg\rvert_{t}.$ That is, if we let $V(t)$ denote the velocity at time $t,$ then their expression $$\frac{dx}{dt}-v$$ is meant to be interpreted as $V(t)-V(0)=V(t)-v.$

So in all, $\frac{dx}{dt}\bigg\rvert^{t}_{0}=\frac{dx}{dt}-v$ by definition of the terms involved. The notation on the left-hand side means "evaluate this at $t$ and at $0$ and subtract", and the notation on the right-hand side is exactly that.