First slide

Evaluation of definite integrals

Question

If $f (x) = \frac{1}{2^{n}}$ when $\frac{1}{2^{n + 1}} < x \leq \frac{1}{2^{n}}, n = 0, 1, 2 \dots$
then $l t_{n \to \infty} \int_{1 / 2^{n}}^{1} f (x) d x$

Moderate

A

$\frac{1}{3}$
B

$\frac{2}{3}$
C

0
D

$\frac{1}{2}$

Solution

$\begin{array}{l} \int_{1 / 2^{n}}^{1} f (x) d x \\ = \int_{1 / 2^{n}}^{1 / 2^{n - 1}} f (x) d x + \int_{1 / 2^{n - 1}}^{1 / 2^{n - 2}} f (x) d x + \dots + \int_{1 / 2}^{1} f (x) d x \end{array}$
$= \frac{1}{2^{n - 1}} [\frac{1}{2^{n - 1}} - \frac{1}{2^{n}}] + \frac{1}{2^{n - 2}} [\frac{1}{2^{n - 2}} - \frac{1}{2^{n - 1}}] + \dots + 1 [1 - \frac{1}{2}]$
$= \frac{1}{2^{2 n - 1}} + \frac{1}{2^{2 n - 3}} + \dots + \frac{1}{2} = \frac{1}{2} [\frac{1 - {(1 / 2^{2})}^{n}}{1 - 1 / 2^{2}}] = \frac{2}{3} [1 - {(\frac{1}{4})}^{n}]$
$lim_{n \to \infty} \int_{1 / 2^{n}}^{1} f (x) d x = \frac{2}{3}$

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