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Comments on Isn't it wrong to write that Indefinite Integral = Definite Integral with a variable in its Upper Limit?

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Isn't it wrong to write that Indefinite Integral = Definite Integral with a variable in its Upper Limit?

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$\int f(t) \; dt = \int_{t_0}^t f(s) \; ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}$

Isn't this wrong? Because LHS $\neq$ RHS in general!

Rather, LHS $\ni$ RHS, because LHS = RHS only if $C = -g(t_0)$.

By the Fundamental Theorem of Calculus (FTC), LHS = $\int{f(t) \; dt} = \{ g(t) + C : C\in \mathbb{R}\}$. It's wrong to write $g(t) + C$ because this means THE antiderivative of $f$. But the LHS is the SET of antiderivatives of $f$ that differ from $g$ by $C$.

By the FTC again, RHS = $\int_{t_0}^t f(s)\ ds = g(t) - g(t_0)$. Then LHS = RHS $\implies C = -g(t_0)$.

Boyce, Elementary Differential Equations (2017 11 edn), pp 28–9. The top-most equation hails from equating (33), but I defined $f(t) :=u(t)g(t)$ and equated (33) to the RHS of (32).

Boyce, Elementary Differential Equations (2017 11 edn), pp 28–9.

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${\int{f(t) \; dt} = \int_{t_0}^t f(s) \; ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}}$

isn't actually in the source text at all. Unpacking some of the surrounding text to more formal notation, it goes from equation (32) $$\exists c: \mu(t) y = \int \mu(t) g(t) \,dt + c$$ so equation (33) $$\exists t_0, c: y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \,ds + c \right)$$

You could argue that it would have been slightly clearer to use $c'$ instead of $c$ in (33), but equally you could argue that it's valid to eliminate $c$ entirely and say that the general solution (my emphasis) is $$y = \frac{1}{\mu(t)} \left( \int_{t_0}^t \mu(s) g(s) \,ds\right)$$ because the arbitrary choice of $t_0$ induces an arbitrary constant of integration. (Actually that would be a bit iffy, because the entire range of possible $t_0$ might not produce a sufficiently wide choice of the constant of integration, but the main point I'm trying to make is that you're not giving enough weight to the context).

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TextKit‭ wrote about 3 years ago

Thanks. You're correct that the equality in question isn't in the text. But to clarify, are you implying that I was wrong to equate (32) with (33)? I explained how I did so in the last para. in my question, above the scan.

Peter Taylor‭ wrote about 3 years ago

@TechnologicallyIlliterate, yes. The wording around arbitrary constants and families of solutions indicates that you need to be more careful than just eliminating the common $y$. The $c$ is (32) is not necessarily equal to the $c$ in (33).