13–20. Mass of one-dimensional objects Find the mass of the following thin bars with the given density function.


ρ(x) = 5e^-2x,for 0≤x≤4

Verified step by step guidance

1. Identify the density function given as \(\rho(x) = 5e^{-2x}\), which represents the mass per unit length of the thin bar over the interval \$0 \leq x \leq 4$.
2. Recall that the total mass \(M\) of a one-dimensional object with density function \(\rho(x)\) over the interval \([a,b]\) is found by integrating the density function over that interval: \(M = \int_{a}^{b} \rho(x) \, dx\).
3. Set up the integral for the mass using the given density and interval: \(M = \int_{0}^{4} 5e^{-2x} \, dx\).
4. To solve the integral, recognize that the integral of \(e^{ax}\) is \(\frac{1}{a} e^{ax}\), so apply this rule with \(a = -2\) and include the constant 5 outside the integral.
5. Evaluate the definite integral by substituting the limits \(x=4\) and \(x=0\) into the antiderivative, then subtract the results to find the total mass.

Key Concepts

Density Function

The density function ρ(x) represents the mass per unit length at a point x along the bar. It varies with position, indicating how mass is distributed along the object. Understanding this function is essential to calculate the total mass by integrating over the given interval.

Definite Integral for Mass

The total mass of a one-dimensional object with variable density is found by integrating the density function over the object's length. Specifically, mass = ∫ from a to b of ρ(x) dx, where [a, b] is the interval along the bar. This integral sums the infinitesimal masses along the bar.

Exponential Functions in Integration

The given density function involves an exponential term e^(-2x). Integrating exponential functions requires knowledge of their antiderivatives, often involving a constant factor adjustment. Recognizing how to integrate e^(kx) is crucial to solving the mass integral correctly.