If f(x)=12n,12n+1<x<12n,n=0,1,2......... and f(0)=0 then ∫01f(x)dx=___

∫12n1f(x)dx=∫21n12n−112n−1dx+∫12n−112n−212n−2dx+......+∫121dx=23(1−14n)∫01f(x)dx=limn→∞∫12n1f(x)dx=23=2∫01x2dx=2limn→∞1n∑r=1n(rn)2

If f(x)=12n,12n+1<x<12n,n=0,1,2......... and f(0)=0 then ∫01f(x)dx=___

∫12n1f(x)dx=∫21n12n−112n−1dx+∫12n−112n−212n−2dx+......+∫121dx=23(1−14n)∫01f(x)dx=limn→∞∫12n1f(x)dx=23=2∫01x2dx=2limn→∞1n∑r=1n(rn)2

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If $f (x) = \frac{1}{2^{n}}, \frac{1}{2^{n + 1}} < x < \frac{1}{2^{n}}, n = 0, 1, 2.........$ and $f (0) = 0$ then $\int_{0}^{1} f (x) d x =___$

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$\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n} {(\frac{r}{n})}^{2}$

$\lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{n} {(\frac{r}{n})}^{3}$

$\lim_{n \to \infty} \frac{2}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{2}$

$\lim_{n \to \infty} \frac{1}{2 n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{3}$

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Detailed Solution

$\int_{\frac{1}{2^{n}}}^{1} f (x) d x = \int_{2^{\frac{1}{n}}}^{\frac{1}{2^{n - 1}}} \frac{1}{2^{n - 1}} d x + \int_{\frac{1}{2^{n - 1}}}^{\frac{1}{2^{n - 2}}} \frac{1}{2^{n - 2}} d x + ...... + \int_{\frac{1}{2}}^{1} d x = \frac{2}{3} (1 - \frac{1}{4^{n}})$

$\int_{0}^{1} f (x) d x = \lim_{n \to \infty} \int_{\frac{1}{2^{n}}}^{1} f (x) d x = \frac{2}{3} = 2 \int_{0}^{1} x^{2} d x = 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{2}$