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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 $l t_{n \to \infty} \int_{1 / 2^{n}}^{1} f (x) d x$

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detailed solution

Correct option is B

∫1/2n1 f(x)dx=∫1/2n1/2n−1 f(x)dx+∫1/2n−11/2n−2 f(x)dx+⋯+∫1/21 f(x)dx=12n−112n−1−12n+12n−212n−2−12n−1+⋯+11−12=122n−1+122n−3+⋯+12=121−1/22n1−1/22=231−14nlimn→∞ ∫1/2n1 f(x)dx=23

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If f(x)=12nwhen 12n+1<x≤12n,n=0,1,2…then ltn→∞∫1/2n1 f(x)dx

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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 $l t_{n \to \infty} \int_{1 / 2^{n}}^{1} f (x) d x$