The value of the lim n -01 x10sin n x d x equals

I_N=[ cos (n x) x^10n]_0^1+10 ∫_0^1cos (n x) x^9 d xn =0+10n[[sin (n x) x^9n]_0^1 9n∫_0^1sin (n x) x^8 d xn] = nbsp; 10 × 9n^2[∫_0^1sin (n x) x^8 d x] = nbs ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

The value of ...

Question

The value of the $lim n \to \infty \int_{0}^{1} x^{10} sin (n x) d x$ equals

$0$

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$\frac{1}{10!}$

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$\frac{π}{2}$

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$1$

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Solution

The correct option is A
$0$

$I_{N} = {[\frac{- cos (n x) x^{10}}{n}]}_{0}^{1} + 10 \int_{0}^{1} \frac{cos (n x) x^{9} d x}{n}$
$= 0 + \frac{10}{n} ⎡ ⎣ {[\frac{sin (n x) x^{9}}{n}]}_{0}^{1} - \frac{9}{n} \int_{0}^{1} \frac{sin (n x) x^{8} d x}{n} ⎤ ⎦$
$= - \frac{10 \times 9}{n^{2}} [\int_{0}^{1} sin (n x) x^{8} d x]$
$= \frac{10!}{n^{10}} [\int_{0}^{1} sin (n x) d x]$
$= 0 as Denominator ⟶ \infty$

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The value of the limn→∞∫10x10sin(nx)dx equals

The correct option is A 0 IN=[−cos(nx)x10n]10+10∫10cos(nx)x9dxn =0+10n⎡⎣[sin(nx)x9n]10−9n∫10sin(nx)x8dxn⎤⎦ =−10×9n2[∫10sin(nx)x8dx] =10!n10[∫10sin(nx)dx] =0 as Denominator ⟶∞

The value of the $lim n \to \infty \int_{0}^{1} x^{10} sin (n x) d x$ equals