Question

Please help me with this question, I don't know where to exactly start

Answer 1

3.

A.
find the area under the curve first
$\int\limits^{\sqrt[3]{\pi}}_0 {x sin(x^3)} \, dx =0.61157809233184$
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
$\int\limits^m_0 {mx-x^2} \, dx =8$
solve
$\int\limits^m_0 {mx-x^2} \, dx =[\frac{mx^2}{2}-\frac{x^3}{3}]^m_0$  $(\frac{m(m)^2}{2}-\frac{m^3}{3})-(0)=\frac{m^3}{2}-\frac{m^3}{3}=\frac{m^3}{6}=8$

so
$\frac{m^3}{6}=8$
m³=48
m=2∛6