Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.7.21

Chapter 6, Problem 6.7.21

Work from force How much work is required to move an object from x=0 to x=3 (measured in meters) in the presence of a force (in N) given by F(x)=2x acting along the x-axis?

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Identify the given force function: $F(x) = 2x$, which varies with position $x$ along the x-axis.

Recall that work done by a variable force along a straight line from $x=a$ to $x=b$ is given by the definite integral: $W = \int_{a}^{b} F(x) \, dx$.

Set up the integral for the work done moving the object from $x=0$ to $x=3$: $W = \int_{0}^{3} 2x \, dx$.

Integrate the function \$2x$ with respect to $x$: find the antiderivative of $2x$, which is $x^2$.

Evaluate the definite integral by substituting the limits $x=3$ and $x=0$ into the antiderivative and subtracting: $W = [x^2]_{0}^{3} = 3^2 - 0^2$.

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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. Instead of using W = F × d, we use the integral W = ∫ F(x) dx from the initial to the final position to account for changes in force.

Definite Integral in Calculus

A definite integral computes the accumulation of quantities, such as area under a curve, over a specific interval. In this context, integrating F(x) from x=0 to x=3 sums the incremental work done at each point along the path.

Force as a Function of Position

When force depends on position, it is expressed as F(x). Understanding how to interpret and manipulate such functions is essential for setting up the integral correctly and solving for work done over a distance.

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