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 =[(mx^2)/(2)-(x^3)/(3)]^m_0`
`((m(m)^2)/(2)-(m^3)/(3))-(0)=(m^3)/(2)-(m^3)/(3)=(m^3)/(6)=8`

so

`(m^3)/(6)=8`
m³=48
m=2∛6