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#3: Post edited
by
The Amplitwist
·
2021-02-02T21:29:00Z (almost 4 years ago)
fixed some MathJax markup
Isn't it wrong to write that Indefinite Integral = Definite Integral with a variable in its Upper Limit?
>${\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 $i$ RHS, because LSH = 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).
- >$\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 $
- i$ 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).
- 
#2: Post edited
by
TextKit
·
2021-01-31T23:54:31Z (almost 4 years ago)
How can an Indefinite Integral = Definite Integral with a variable in its Upper Limit?
- Isn't it wrong to write that Indefinite Integral = Definite Integral with a variable in its Upper Limit?
#1: Initial revision
by
TextKit
·
2021-01-31T23:53:38Z (almost 4 years ago)
How can an Indefinite Integral = Definite Integral with a variable in its Upper Limit?
>${\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 LSH = 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).
