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I read the derivation of $v=u+at$ using integration. The steps are as follows -

$$\frac{dv}{dt}=a dt$$ $$dv=adt$$ $$\int dv=\int a dt$$ $$\int dv=a\int dt$$ $$v=at+c$$ My questions are as follows -

1. On integrating $dt$ on the RHS we get a $+c$ (constant of integration) but why is there no $+c$ on the LHS while integrating $dv$?

2.Can we take out something of the integration symbol as $a$ is taken out is the second last step?

3.Why in the the last step it is $at+c$ and not $at+ac$?

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2 Answers 2

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1) On integrating dt on the RHS we get a +c(constant of integration) but why is there no +c on the LHS while integrating dv?

If we start with:

$$ dv = adt $$

and integrate both sides then we can indeed have a constant of integration on both sides:

$$ v + C_1 = at + C_2 $$

but we can just subtract $C_1$ from both sides to get:

$$ v = at + (C_2 - C_1) = at + C $$

where $C = C_2 - C_1$. So we only need to write a constant of integration on one side.

2) Can we take out something of the integration symbol as a is taken out is the second last step?

I assume you mean "Why can we take etc". The answer is that $a$ is a constant and doesn't depend on $t$ or $x$. If we are integrating a product of two terms, like $a \times t$, then we have to do it using integration by parts. If you look at the equation at the top of the Wikipedia article you'll see that:

$$ \int at = a\int t - \int t \frac{da}{dt} $$

But since $a$ is a constant $da/dt = 0$ and the second term on the right hand side is zero. That means the equation simplifies to;

$$ \int at = a\int t $$

In everyday life we wouldn't bother going through the integration by parts. We just know from experience that you can take constants outside the integral. NB if the acceleation is not constant, i.e. it's a function of time $a(t)$, then $da/dt \ne 0$ we cannot take it outside the integral and that tends to make the integration harder to do.

3) Why in the the last step it is at+c and not at+ac?

Suppose we had:

$$ v = at + Ct $$

Differentiating should give us back what we started with, but differentiating the above equation gives:

$$ \frac{dv}{dt} = a + C $$

which is wrong. On the other hand differentiating $v = at + C$ does give us back $dv/dt = a$.

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In the first step ( $\int dv=a\int dt$), we get $v+c_1=a(t+c_2)$. and therefore $v=at+(ac_2-c_1)$. The term $ac_2-c_1$ is constant (because $a$ ,here, is constant). Instead of tediously writing the integration constants for every step, we can write one at the ending, since the last constant ($c$, in this case) absorbs all the other constants which have turned up.

Also, you can take things out of an integral as long as they are constant.

(This page lists out some basic properties of integrals: http://en.wikibooks.org/wiki/Calculus/Indefinite_integral#Basic_Properties_of_Indefinite_Integrals )

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