Work from force How much work is required to move an object from x=1 to x=3 (measured in meters) in the presence of a force (in N) given by F(x) = 2x² acting along the x-axis?

Verified step by step guidance

1. Identify that the work done by a variable force along the x-axis from position \(x=a\) to \(x=b\) is given by the definite integral of the force function: \(W = \int_{a}^{b} F(x) \, dx\).
2. Substitute the given force function \(F(x) = 2x^{2}\) and the limits of integration \(a=1\) and \(b=3\) into the integral: \(W = \int_{1}^{3} 2x^{2} \, dx\).
3. Set up the integral expression explicitly: \(W = 2 \int_{1}^{3} x^{2} \, dx\) to simplify the integration process.
4. Recall the integral formula for a power function: \(\int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C\). Use this to find the antiderivative of \(x^{2}\), which is \(\frac{x^{3}}{3}\).
5. Evaluate the definite integral by applying the Fundamental Theorem of Calculus: compute \$2 \left[ \frac{x^{3}}{3} \right]_{1}^{3} = 2 \left( \frac{3^{3}}{3} - \frac{1^{3}}{3} \right)$ to express the work done.

Key Concepts

Work Done by a Variable Force

Work done by a force that varies with position is calculated by integrating the force function over the displacement interval. Specifically, work W = ∫ F(x) dx from the initial to the final position, capturing how the force changes along the path.

Definite Integral in Calculus

A definite integral computes the accumulation of quantities, such as area under a curve or total work done, between two limits. It sums infinitesimal contributions over an interval, providing an exact value for the total effect.

Force as a Function of Position

When force depends on position, it must be expressed as a function F(x). Understanding this relationship allows us to apply calculus techniques to find work or energy, as the force magnitude changes with location along the x-axis.