# Taylor series expansion and evaluating an integral

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.

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

$f(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} +\ ...=\frac{1}{1-x/2}=\frac{2}{2-x}\ for \ x \in[-1, 1].$
$$\sqrt{\exp(\int_{-1}^{1}f(x)dx)}=\sqrt{\exp(-2\ln(2-x)|_{-1}^1)}=\sqrt{\exp(0+2\ln3)}=\sqrt{9}=3$$