NLt- -13-n4-14-23-n4-24-1-2n-

The correct option is D 1/4log 2 nLt→∞∑_r=1^nr^3r^4+n^4=nLt→∞∑_r=1^n1n #10;( (rn)^31+(rn)^4) #10; =∫_0^1x^31+x^4· #160; dx=14log #10;( ...

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Standard XII

Mathematics

Definite Integral as Limit of Sum

nLt→∞ [13/n4+...

Question

$n L t \to \infty [\frac{1^{3}}{n^{4} + 1^{4}} + \frac{2^{3}}{n^{4} + 2^{4}} + \dots + \frac{1}{2 n}] =$

\frac{1}{4} log 4

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\frac{1}{2} log 2

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\frac{1}{4} log 3

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\frac{1}{4} log 2

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Solution

The correct option is D $\frac{1}{4} log 2$
$n L t \to \infty \sum_{r = 1}^{n} \frac{r^{3}}{r^{4} + n^{4}} = n L t \to \infty \sum_{r = 1}^{n} \frac{1}{n} ⎛ ⎜ ⎜ ⎝ \frac{(\frac{r}{n})^{3}}{1 + (\frac{r}{n})^{4}} ⎞ ⎟ ⎟ ⎠$
$= \int_{0}^{1} \frac{x^{3}}{1 + x^{4}} \cdot d x = \frac{1}{4} l o g (1 + x^{4}) \int_{0}^{1}$
$= \frac{1}{4} l o g 2$

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nLt→∞[13n4+14+23n4+24+…+12n]=

The correct option is D 14log2 nLt→∞∑nr=1r3r4+n4=nLt→∞∑nr=11n⎛⎜ ⎜⎝(rn)31+(rn)4⎞⎟ ⎟⎠ =∫10x31+x4⋅dx=14log(1+x4)∫10 =14log2

$n L t \to \infty [\frac{1^{3}}{n^{4} + 1^{4}} + \frac{2^{3}}{n^{4} + 2^{4}} + \dots + \frac{1}{2 n}] =$