Q&A

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

+3
−0

$\int f(t) \; dt = \int_{t_0}^t f(s) \; ds \quad \text{ where$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).

Why does this post require moderator attention?
Why should this post be closed?

You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.

+2
−0

${\int{f(t) \; dt} = \int_{t_0}^t f(s) \; ds \quad \text{ where$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).

Why does this post require moderator attention?

+0
−0

Your observation RHS $\in$ LHS is correct. LHS is not a unique function but a set of functions differing by a constant.

I take issue with your usage of $C$. There is not really a meaningful definition of $C$ for an arbitrary antiderivative. For example, what is the constant term of $\cos(x) + 3$? You could say it's 3 but what if we rewrite it as $u(x) + 4$ where $u(x) = \cos(x) + 1$?

Note that if $g \in \int f$ then $\int_{t_0}^t f = g(t) - g(t_0)$. Whatever constant is included in $g$ cancels in $g(t) - g(t_0)$ so $g$ and $t_0$ don't have anything to do with each other.

If instead we write $F = \int_{t_0}^t f$ then we have $F(t_0) = \int_{t_0}^{t_0} t = 0$. This equation forces a certain constant term of $F$. A different choice of $t_0$ can (but doesn't necessarily) lead to a different constant term of $F$.

Why does this post require moderator attention?

This community is part of the Codidact network. We have other communities too — take a look!

Want to advertise this community? Use our templates!

Like what we're doing? Support us!