The area of ​​the lines y = x2 and y = m finite shape is 36. Find the value of the parameter m.

Answer 1

integrate m-x^2 = mx -x^3/3 where the line and curve intersect

y = m = x^2

x = sqr(m)

m(sqr(m)) - (sqr(m)^3/3 = m^(3/2) = (1/3)m^(3/2) = (2/3)m^(3/2) = 36

m^(3/2) = 36(3/2) = 18(3) = 54

m^3 = 54^2 = 2916

m = cube root of 2916 = about 14.29

area between y=about 14.29 and y=x^2 is 36

Answer 2

x^2=m --> x=±(m)^(1/2) points where x^2 intersects with real line (OX axis)

36=∫_-√m^√mx²dx= 2∫₀^√m x²dx =(2/3)m^(3/2)

So 36=(2/3)m^(3/2)

Now solve for m

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