Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.2.15

Chapter 6, Problem 6.2.15

Determine the area of the shaded region in the following figures.

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Identify the curves that bound the shaded region. Here, the region is bounded by the curves \(y = \sin x\) and \(y = \cos x\) between \(x = 0\) and \(x = \frac{\pi}{2}\).

Determine which function is on top and which is on the bottom in the interval \([0, \frac{\pi}{2}]\). Since \(\cos x\) starts at 1 and decreases, and \(\sin x\) starts at 0 and increases, \(\cos x\) is above \(\sin x\) in this interval.

Set up the integral for the area of the shaded region as the integral of the difference between the top function and the bottom function over the interval: \(\int_0^{\frac{\pi}{2}} (\cos x - \sin x) \, dx\).

Integrate the function \(\cos x - \sin x\) with respect to \(x\). Recall that the integral of \(\cos x\) is \(\sin x\) and the integral of \(\sin x\) is \(-\cos x\).

Evaluate the definite integral by substituting the limits \(0\) and \(\frac{\pi}{2}\) into the antiderivative and subtracting to find the exact area of the shaded region.

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Key Concepts

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

Definite Integrals and Area Between Curves

The area between two curves over an interval can be found using definite integrals by integrating the difference of the functions. Specifically, the area is the integral of the upper function minus the lower function from the left to the right boundary of the region.

Trigonometric Functions and Their Properties

Understanding the behavior of sine and cosine functions, including their values and intersections within the interval [0, π/2], is essential. Knowing where sin x and cos x intersect helps determine the limits and which function is on top in the region.

Setting Up the Integral Limits

Identifying the correct interval for integration is crucial. Here, the shaded region lies between x = 0 and x = π/2, where the two curves intersect and bound the area. Properly setting these limits ensures accurate calculation of the shaded area.

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Related Practice

Textbook Question

53–62. Choose your method Let R be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when R is revolved about the given axis.

y = x−x⁴,y=0; about the x-axis.

Textbook Question

52–54. Force on a window A diving pool that is 4 m deep and full of water has a viewing window on one of its vertical walls. Find the force on the following windows.

The window is a square, 0.5 m on a side, with the lower edge of the window on the bottom of the pool.

Textbook Question

Find the area of the surface generated when the given curve is revolved about the given axis.

y=1/4(e^2x+e^−2x), for −2≤x≤2; about the x-axis

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Textbook Question

Find the area of the surface generated when the given curve is revolved about the given axis.

y=√1−x^2, for −1/2≤x≤1/2; about the x-axis

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Textbook Question

9-34. Shell method Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about indicated axis.

{Use of Tech} y = In x/x²,y = 0,x = 3, about the y-axis

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Textbook Question

Determine the area of the shaded region in the following figures.