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Let $$ f(x) \leq g(x), \forall\, x.\hspace{0.5cm} (1)$$

Moreover, considering the indefinite integrals

$$\int f(x)\,dx= F(x) + C_1 \hbox{ and } \int g(x)\,dx = G(x) + C_2.$$

My question: If we supppose (1), is true that $$ F(x) + C_1=\int f(x)\,dx \leq \int g(x)\,dx = G(x) + C_2? $$

If the answer is wrong, what would be a counterexample or what the inconsistency in this it?

Welljc
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  • No, take $g=f = 0$. Then, since you have indefinite integrals, $\int f = 0+c_1, \int g =0+c_2$, but the $c_k$ are unspecified constants, so it can't be true unless you require that $c_1 \le c_2$. – copper.hat Oct 30 '14 at 15:16

2 Answers2

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Clearly false: $C_1$ and $C_2$ are as large as you like. What is true is that $$ \int_a^x f(t)\, dt \leq \int_a^x g(t)\, dt $$ for any $x$. In other words, you must take $C_1=C_2$.

Siminore
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Observe that $f(x) \leq g(x)$ is a statement about a single function $f$ and a single function $g$, evaluated pointwise. On the other hand, $\int f(x) \, dx \leq \int g(x) \, dx$ is about a family of functions.

What do you mean when you say one family of functions is larger than another? If you mean that any selection from the antiderviatives of $f$ is smaller than any antiderivative of $g$, you are incorrect, for example take $f(x) = g(x)$ and $C_1 > C_2$.

Jason Knapp
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