2
$\begingroup$

I have the following type of inequality: $$ \int_0^{h(x)}\mathrm e^{f(t)} g(t) \mathrm dt> \int_0^{h(x)}\mathrm e^{f(t)} f(t) \mathrm dt $$ Question:

Can I cancel the term $\mathrm e^{f(t)}$, as it appears on both sides and the limits of integration are equal?

Thanks a lot

$\endgroup$

closed as off-topic by Xander Henderson, mrtaurho, Cesareo, Paul Frost, metamorphy Jul 1 at 10:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, mrtaurho, Cesareo, Paul Frost, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 8
    $\begingroup$ No, you can't. What makes you think you could do it? $\endgroup$ – Jakobian Jun 28 at 12:50
  • $\begingroup$ This question is not off-topic $\endgroup$ – David Jul 5 at 8:50
5
$\begingroup$

No, you can't. Although, you can claim that $$ \int_{0}^{h(x)} e^{f(t)}(g(t) - f(t)) dt > 0, $$ which is indeed not the same as $$ \int_{0}^{h(x)} g(t) - f(t) dt > 0. $$ You can think of $e^{f(t)}$ as a weight-function.

$\endgroup$
  • $\begingroup$ Thanks, seeing it like this makes it clear.... $\endgroup$ – HJ Creens Jun 28 at 12:55
1
$\begingroup$

You can't. Check by yourself with the following counterexample

Let $f(t)=t; g(t)=3.1$ and $h(x)=5$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.