Just 3 Steps

Calculus Level 4

Evaluate :

$\lim_{n \to \infty } n^{2} \int_{\frac{-1}{n}}^{\frac{1}{n}} ( 2007\sin x + 2008 \cos x) |x| dx$

The answer is 2008.

2 solutions

Krishna Sharma
Jan 5, 2015

Rewriting above equation

$\lim_{n \to \infty } \dfrac{\int_{\frac{-1}{n}}^{\frac{1}{n}} (2007sinx + 2008cosx)|x| dx}{\frac{1}{n^{2}}}$

Note we can apply Newton-Leibnitz Theorem $(\frac{0}{0} form)$

$\lim_{n \to \infty } \dfrac{(2007sin(\frac{1}{n}) + 2008cos(\frac{1}{n}))(\frac{-1}{n^{3}}) - (2007sin(\frac{-1}{n}) + 2008cos(\frac{-1}{n}))(\frac{1}{n^{3}}))}{\frac{-2}{n^{3}}}$

Now after simplifing $\sin(\frac{1}{n})$ & $n^{3}$ terms will cancel out, finally we will get

$\lim_{n \to \infty } \dfrac{4016cos(\frac{1}{n})}{2} = \boxed{2008}$

$\sin x$ is an odd function, so you can just get rid of it.

Jake Lai - 6 years, 4 months ago

Shaikh Waz Noori
Mar 18, 2015

I first simplified mod(X) then used the integration (by dividing the integral to two parts -1/n to 0 and 0 to 1/n ) then used the limits to get the value ....that was simple....

one could have used another method ie.. firstly eleminate mod x sinx part as it is an odd function which on symmetric domain of x gives the value of integral as 0. now we could use the series expansion of cosx in neighbourhood of 0 and neglected the higher order terms of 1/n .It striked me first.then i figured other methods tooo

Anmol Agarwal - 4 years, 5 months ago

Just 3 Steps

The answer is 2008.

2 solutions

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