# Cancel out of integrals [closed]

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

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

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• No, you can't. What makes you think you could do it? – Jakobian Jun 28 at 12:50
• This question is not off-topic – 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.
Let $$f(t)=t; g(t)=3.1$$ and $$h(x)=5$$