Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.5.38

Chapter 6, Problem 6.5.38

Function defined as an integral Write the integral that gives the length of the curve y = f(x) = ∫₀^x sin t dt on the interval [0,π]

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Recall the formula for the length of a curve defined by a function \(y = f(x)\) on the interval \([a,b]\): the length \(L\) is given by the integral \(L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\).

Identify the function \(f(x)\) given as an integral: \(f(x) = \int_0^x \sin t \, dt\).

Find the derivative \(\frac{dy}{dx}\) of the function \(f(x)\) using the Fundamental Theorem of Calculus, which states that if \(f(x) = \int_a^x g(t) \, dt\), then \(f'(x) = g(x)\). So, \(\frac{dy}{dx} = \sin x\).

Substitute \(\frac{dy}{dx} = \sin x\) into the length formula to get \(L = \int_0^{\pi} \sqrt{1 + (\sin x)^2} \, dx\).

Write the final integral expression for the length of the curve on \([0, \pi]\) as \(L = \int_0^{\pi} \sqrt{1 + \sin^2 x} \, dx\).

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Key Concepts

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

Arc Length Formula

The arc length of a curve y = f(x) from x = a to x = b is given by the integral ∫_a^b √(1 + (dy/dx)²) dx. This formula calculates the length by summing infinitesimal line segments along the curve, accounting for both horizontal and vertical changes.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if f(x) = ∫_a^x g(t) dt, then f'(x) = g(x). It allows us to find the derivative of an integral-defined function, which is essential for computing dy/dx in the arc length formula.

Derivative of the Given Function

Given y = ∫₀^x sin t dt, by the Fundamental Theorem of Calculus, dy/dx = sin x. Knowing this derivative is crucial to substitute into the arc length formula to express the integral that calculates the curve's length on [0, π].

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