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How can I solve this:

Let $$f(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} +\ldots \text{ for } x \in[-1, 1]. $$ Evaluate: $$ \sqrt{\exp\left(\int_{-1}^{1}f(x)\;dx\right)} $$

I think I need to use Taylor series expansion for the function here, but I am really stuck on how to do it.

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  • $\begingroup$ Do you know how to compute $f(x)$ explicitly? E.g., recognizing $\sum_{n=0}^\infty \frac{x^n}{2^n}$. $\endgroup$ – Clement C. Jun 13 '17 at 13:54
  • $\begingroup$ $f(x)$ is a geometric series with common ratio $x/2$. $\endgroup$ – Zubzub Jun 13 '17 at 13:56
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$f(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} +\ ...=\frac{1}{1-x/2}=\frac{2}{2-x}\ for \ x \in[-1, 1].$

$$\sqrt{\exp(\int_{-1}^{1}f(x)dx)}=\sqrt{\exp(-2\ln(2-x)|_{-1}^1)}=\sqrt{\exp(0+2\ln3)}=\sqrt{9}=3$$

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