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What is the net area between #f(x) = cos2x-xsinx # and the x-axis over #x in [0, 3pi ]#?

Answer 1

Ezekiel Alexander

The net area between #f(x)# and the #x#-axis from #0# to #3pi# is #-3pi#.

#color(white)=int_0^(3pi)(cos2x−xsinx)dx#

#=int_0^(3pi)cos2x# #dx−int_0^(3pi)xsinx# #dx#

#=1/2sin2x|_0^(3pi)−int_0^(3pi)xsinx# #dx#

#=1/2sin(3pi)-1/2sin(0)−int_0^(3pi)xsinx# #dx#

#=−int_0^(3pi)xsinx# #dx#

Using the DI method (an easier way to understand integration by parts):

Multiply the diagonals with the appropriate sign, and add them (and don't forget the negative sign in front of the integral from before):

#=-((x*-cosx-1*-sinx)|_0^(3pi))#

#=-((-xcosx+sinx)|_0^(3pi))#

#=-((sinx-xcosx)|_0^(3pi))#

#=-((sin3pi-3picos3pi)-(sin0-0cos0))#

#=-((0+3pi)-(0-0))#

#=-3pi#

That is the net area under the curve. Hope this helped!

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Answer 2

Emilia Aguilar

To find the net area between ( f(x) = \cos(2x) - x\sin(x) ) and the x-axis over ( x ) in ( [0, 3\pi] ), we need to compute the definite integral of the absolute value of ( f(x) ) over the given interval.

( \text{Net Area} = \int_{0}^{3\pi} |f(x)| , dx )

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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.