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?

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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: \(W = \int_{a}^{b} F(x) \, dx\).

Substitute the given force function \(F(x) = 2x^{2}\) and the limits of integration \(a=1\) and \(b=3\) into the integral: \(W = \int_{1}^{3} 2x^{2} \, dx\).

Set up the integral expression explicitly: \(W = 2 \int_{1}^{3} x^{2} \, dx\) to simplify the integration process.

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 \(\frac{x^{3}}{3}\).

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.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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.

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