How do you find the definite integral for: #2(pi)x(cos^(-1)(x))dx# for the intervals #[0, 1]#?

Answer 1

#I=pi^2/4#

Here,

#I=2piint_0^1 xcos^-1x#
Subst. #cos^-1x=u=>x=cosu=>dx=-sinudu#
#x=0=>cosu=0=>u=pi/2and#
# x=1=>cosu=1=>u=0#

So,

#I=2piint_(pi/2)^0 cosu*u(-sinu)du#
#=2piint_0^(pi/2) usinucosudu.to[becauseint_a^bf(x)dx=-int_b^af(x)dx]#
#=piint_0^(pi/2)usin2udu...to[becausesin2theta=2sinthetacostheta]#
#"Using "color(red)"Integratio by Parts :"#
#color(blue)(intu*vdx=u*intvdx-int(u'intvdx)dx#
Take, #u=u andv=sin2u=>u'=1 and intvdu=-(cos2u)/2#
#I=pi{[u*(-cos2u)/2]_0^(pi/2)-int_0^(pi/2)(-cos2u)/2du}#
#=pi{[pi/2*(-cospi)/2-0]+1/2[(sin2u)/2]_0^(pi/2)}#
#=pi{pi/2(1/2)+1/4[sinpi-sin0]}#
#=pi{pi/4}#
#=pi^2/4#
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Answer 2

To find the definite integral of (2\pi x \arccos(x)) over the interval ([0, 1]), you integrate the function with respect to (x) and then evaluate the result at the upper and lower bounds of the interval, and subtract the lower result from the upper result.

The antiderivative of (2\pi x \arccos(x)) is (x^2 \arccos(x) + \frac{\pi}{2}x^2 - \frac{x^3}{3}).

Evaluating this antiderivative at the upper limit (1) yields:

[1^2 \arccos(1) + \frac{\pi}{2} \times 1^2 - \frac{1^3}{3}]

[= 0 + \frac{\pi}{2} - \frac{1}{3}]

Evaluating at the lower limit (0) yields:

[0^2 \arccos(0) + \frac{\pi}{2} \times 0^2 - \frac{0^3}{3}]

[= 0]

Therefore, the definite integral is:

[\left(\frac{\pi}{2} - \frac{1}{3}\right) - 0 = \frac{\pi}{2} - \frac{1}{3}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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