What is the net area between #f(x) = x^2-ln(x^2+1) # and the x-axis over #x in [1, 2 ]#?

Answer 1

#=1/3-ln(5^2/2)+2arctan2-pi/2#

To find the area between #f(x)# and x-axis is to find the definite integral of this function between #[1,2]#
#int_1^2f(x)#
#=int_1^2x^2-ln(x^2+1)dx# #=color(brown)(int_1^2x^2dx)-color(blue)(int_1^2ln(x^2+1)dx)#
Let us compute the first integral #color(brown)(int_1^2x^2dx# As we know integral of the polynomial #intx^n=x^(n+1)/(n+1)# So #color(brown)(intx^2dx=x^3/3)#
#color(brown)(int_1^2x^2dx=2^3/3-1^3/3=8/3-1/3=7/3#
Now, Let us compute the second integral #color(blue)(int_1^2ln(x^2+1)dx)# using integration by parts .
Let #color(red)(u(x)=lnx^2+1)rArrdcolor(|red)(u(x)=(2x)/(x^2+1)dx)# then #color(purple)(dv=dx)rArrcolor(purple)(v(x)=x)#
#color(blue)(intln(x^2+1)dx)# #=color(red)(u(x))*color(purple)(v(x))-intcolor(purple)(v(x))*color(red)(du(x))#
#intcolor(purple)(v(x))*color(red)(du(x))#
#=int x*(2x)/(x^2+1)dx#
#=int (2x^2)/(x^2+1)dx#
#=2int(x^2)/(x^2+1)dx# #=2int1-1/(x^2+1)dx# #=2intdx-2int1/(x^2+1)dx#
#=color(orange)(2x-2arctanx)#
So, #color(blue)(intln(x^2+1)dx)# #=color(red)(u(x))*color(purple)(v(x))-intcolor(purple)(v(x))*color(red)(du(x))#
#=color(purple)xcolor(red)(ln(x^2+1))-(color(orange)(2x-2arctanx))#
#=xln(x^2+1)-2x+2arctanx#
#color(blue)(int_1^2ln(x^2+1)dx)# #=2ln(2^2+1)-2*2+2arctan2-(1*ln(1^2+1)-2*1+2arctan1)#
#=2ln5-4+2arctan2-(ln2-2+2arctan1)# #=2ln5-4+2arctan2-ln2+2-2*pi/4# #=2ln5-ln2-4+2+2arctan2-pi/2# #=ln(5^2)-ln2-2+2arctan2-pi/2# #color(blue)(=ln(5^2/2)-2+2arctan2-pi/2#
#int_1^2f(x)#
#=int_1^2x^2-ln(x^2+1)dx# #=color(brown)(int_1^2x^2dx)-color(blue)(int_1^2ln(x^2+1)dx)# #=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2# #=color(brown)(7/3)-color(blue)(ln(5^2/2)-2+2arctan2-pi/2# #=7/3-2-ln(5^2/2)+2arctan2-pi/2# #=7/3-6/3-ln(5^2/2)+2arctan2-pi/2#
#=1/3-ln(5^2/2)+2arctan2-pi/2#
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Answer 2

To find the net area between ( f(x) = x^2 - \ln(x^2 + 1) ) and the x-axis over ( x ) in the interval ([1, 2]), we need to compute the definite integral of ( f(x) ) from 1 to 2.

[ \text{Net Area} = \int_{1}^{2} (x^2 - \ln(x^2 + 1)) , dx ]

After integrating, the net area will be the result.

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Answer 3

To find the net area between ( f(x) = x^2 - \ln(x^2 + 1) ) and the x-axis over ( x ) in the interval ([1, 2]), we need to evaluate the definite integral of ( |f(x)| ) over the given interval.

[ \text{Net area} = \int_{1}^{2} |f(x)| , dx ]

First, we need to determine the intervals where ( f(x) ) is above or below the x-axis over the interval ([1, 2]). We find this by examining the sign of ( f(x) ) for ( x ) in the interval.

Then, we calculate ( |f(x)| ) for each interval and integrate over those intervals separately. The net area will be the sum of the areas above the x-axis minus the areas below the x-axis.

Finally, we compute the definite integral of ( |f(x)| ) over each interval using appropriate integration techniques. The resulting net area will be the solution to the problem.

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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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