3.
A.
find the area under the curve first
[tex] \int\limits^{\sqrt[3]{\pi}}_0 {x sin(x^3)} \, dx =0.61157809233184[/tex]
then solve c∛π=0.61157809233184
c≈0.41757577488026
round if necicary
4.
alright, solve where they itnersect
x²=y=mx
x²=mx
x²-mx=0
x(x-m)=0
set to zero
x=0
x-m=0
x=m
they intersect at x=0 and x=m
which one is on top?
hmm
y=mx will always be on top (I don't know how to prove that but it's on top)
so we do
[tex] \int\limits^m_0 {mx-x^2} \, dx =8[/tex]
solve
[tex] \int\limits^m_0 {mx-x^2} \, dx =[\frac{mx^2}{2}-\frac{x^3}{3}]^m_0[/tex] [tex](\frac{m(m)^2}{2}-\frac{m^3}{3})-(0)=\frac{m^3}{2}-\frac{m^3}{3}=\frac{m^3}{6}=8[/tex]
so
[tex]\frac{m^3}{6}=8[/tex]
m³=48
m=2∛6
