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how to calculate integral,

$\displaystyle\lim_{h \to -\frac{1}{2}} \int_0^{1+2h}\frac{\sin(x^2)}{(1+2h)^3}dx$

Not sure if the limit exists or not.

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3 Answers 3

up vote 3 down vote accepted

We may change your limit by $$\displaystyle\lim_{\varepsilon \to 0}\frac{1}{{\varepsilon^3}} \int_0^{\varepsilon}{\sin(x^2)}dx$$

At this point, you may use L'Hopital's Rule.

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$$ \sim {1 \over \epsilon^{3}}\int_{0}^{\epsilon}x^{2}\,{\rm d}x = \color{#ff0000}{\large{1 \over 3}} $$

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1  
you should be a bit more explicit how you come to this solution –  Dominic Michaelis Oct 14 '13 at 7:08

Since the range of integration is small, replace Sin[x^2] by its Taylor expansion around x=0 or just remember that Sin[x]/x has a limit of 1 when x->0. Are you able to continue with this ?

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