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


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  • 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

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.

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

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

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


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