Use L’Hospital’s rule to evaluate limx21x-2dt2xsinttdt.

Short Answer

Expert verified

The value oflimx21x-2dt2xsinttdt is12sin2

Step by step solution

01

Given Information

Given that limx21x-2dt2xsinttdt …... (1)

02

Finding limx→21x-2dt∫2xsinttdt=6

Use L-Hospital’s rule to differentiate the numerator and denominator separately.

Letddx2xsinttdt

We know thatddxu(x)v(x)f(x,t)dt=f(x,v)dvdx-f(x,u)dudx+uvfxdt…... (2)

Compare equation (1) and equation (2), then the given function becomes

ddx2xsinttdt=sinxx·ddxx-sin22·ddx2+2x0.dtddx2xsinttdt=sinxx1-sin220ddx2xsinttdt=sinxx,andddxx-2=1

Then1x-22xsinttdt=sinxx

Therefore,

limx21x-2dt2xsinttdt=limx2sinxxlimx21x-2dt2xsinttdt=12sin2

Hence, limx21x-2dt2xsinttdt=12sin2.

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