Find the area between the graphs y=2x and y=x^3
Solving the two equations:x(x^2-2)=0 → x=0 ,x=±√2From the fig 1.8, the area is:A=∫_(-√2)^0(x^3-2x)dx+∫_0^(√2)(2x-x^3 ) dx=[1/4 x^4-x^2 ]_(-√2)^0+[x^2-1/4 x^4 ]_0^√2=-(1-2)+(2-1)=2 square units
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Solving the two equations:
need an explanation for this answer? contact us directly to get an explanation for this answerx(x^2-2)=0 → x=0 ,x=±√2
From the fig 1.8, the area is:
A=∫_(-√2)^0(x^3-2x)dx+∫_0^(√2)(2x-x^3 ) dx
=[1/4 x^4-x^2 ]_(-√2)^0+[x^2-1/4 x^4 ]_0^√2
=-(1-2)+(2-1)=2 square units