Integrals Analysis: Solving ∫ [-2pi,2pi] (x^2)(sin(e^x))^8 dx

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SUMMARY

The integral ∫ [-2pi,2pi] (x^2)(sin(e^x))^8 dx is bounded above by (16pi^3)/3. The discussion highlights that the function (sin(e^x))^8 approaches zero for x < 0, allowing for a comparison with the simpler integral ∫ [-2pi,2pi] x^2 dx. By establishing that x^2 is greater than or equal to (sin(e^x))^8, the inequality ∫ f(x) >= ∫ g(x) is confirmed, validating the upper bound.

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Homework Statement


∫ [-2pi,2pi] (x^2)(sin(e^x))^8 dx <= (16pi^3)/3


Homework Equations





The Attempt at a Solution



I know that the function is almost 0 when x<0 I'm just not sure how to show this mathematically
 
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how about considering the simpler integral first
\int_{-2 \pi}^{2 pi} dx (x^2)
 
lanedance said:
how about considering the simpler integral first
\int_{-2 \pi}^{2 pi} dx (x^2)

thanks I was able to figure this out not too long ago.

I just say that x^2 >= the other function and thus ∫ f(x) >= ∫ g(x) on [a,b]
since ∫ x^2 [-2pi,2pi] is equal to (16pi^3)/3 and ∫ f(x) >= ∫ g(x) the inequality holds

thanks though
 

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