Question 1032445
*[illustration fd2.JPG].
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{{{A=int((f(x)),dx,0,1)+int((-f(x)),dx,1,2)}}}
{{{A=int((1-x^2),dx,0,1)+int((x^2-1),dx,1,2)}}}
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{{{A[1]=x-x^3/3+C}}}
{{{A[1]=(1-0)-(1^3-0^3)/3}}}
{{{A[1]=1-1/3}}}
{{{A[1]=2/3}}}
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{{{A[2]=x^3/3-x+C}}}
{{{A[2]=(2^3-1^3)/3-(2-1)}}}
{{{A[2]=(8-1)/3-1}}}
{{{A[2]=7/3-3/3}}}
{{{A[2]=4/3}}}
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{{{A=A[1]+A[2]}}}
{{{A=2/3+4/3}}}
{{{A=2}}}
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 This previous method assumes the area below the x-axis is treated as positive also.
If you treat this as negative then the sum would change to,
{{{A=A[1]-A[2]}}}
{{{A=2/3-4/3}}}
{{{A=-2/3}}}