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lozanna [386]
3 years ago
6

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

Advanced Placement (AP)
1 answer:
Anarel [89]3 years ago
5 0
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
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