# Physics

A long thin rod lies along the x-axis from the origin to x=L, with L= 0.830 m. The mass per unit length, $\lambda$ (in kg/m) varies according to the equation $\lambda = \lambda_0 (1+1.110x^3). The value of$\lambda_0$is 0.200 kg/m and x is in meters. Calculate the total mass of the rod. - So I tried: lambda = m/x m = (lambda)(x) = (0.200kg/m)(0.830m) = 0.166kg ...which was apparently wrong so then I retried this time using the given equation:$ \lambda = \lambda_0 (1+1.110x^3).
= (0.200kg/m)(1+(1.110)(0.830m)^3)
= 0.271kg

But that is also incorrect and I don't know what else to do.

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1. this is integral calculus

the mass of a segment of the rod is
... segment length * lambda

lambda = .200 (1 + 1.110 x^3)
... = .200 + .222 x^3

segment length is dx

integrate from x = 0 to .83

.200 dx + .222 x^3 dx ... sum the two integrals

.166 + .026

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2. I have no idea because I'm not in your grade. SORRY!

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3. dM=dx(0.200*(1+1.110x^3))
taking integral on both sides,
M=0.200(L+1.110*x^4/4)
put L=0.830
M=0.19234

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