A Weird Square Root

Calculus Level 3

Let f ( x ) = 1 + x 2 + x 2 4 + x 3 8 + \displaystyle f(x)= 1 + \frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+\cdots , for 1 x 1 -1\le x \le 1 .

Find e 0 1 f ( x ) d x \large \sqrt{e^{\int_0^1 f(x)\, dx}} .

0 0 ln 2 \ln 2 2 2 1 1

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1 solution

Hana Wehbi
Nov 8, 2016

We know that f ( x ) f(x) is an infinite geometric series, thus the infinite sum is 1 1 x 2 = 2 2 x \frac{1}{1-\frac{x}{2}} = \frac{2}{2-x} .

Thus, 0 1 2 2 x = 2 ln 2 \int_{0}^{1}\frac{2}{2-x} = 2\ln 2 .

Then e 2 ln 2 = 2 2 = 2 \sqrt{e^{2\ln2}}= \sqrt{2^2}= 2 .

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