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Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 5 - IntegralsProblem 5.PE.23
Chapter 5, Problem 5.PE.23

Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.
y = sin x, y = x, 0 ≤ x ≤ π/4

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1
Identify the region enclosed by the curves y = sin x and y = x over the interval 0 \(\leq\) x \(\leq\) \(\frac{\pi}{4}\). Since both functions are continuous and the interval is closed, the area between them can be found by integrating the difference of the functions over this interval.
Determine which function is on top and which is on the bottom in the interval. Compare y = sin x and y = x for 0 \(\leq\) x \(\leq\) \(\frac{\pi}{4}\) to find which one is greater. This will tell us the order of subtraction inside the integral.
Set up the integral for the area A as the integral from 0 to \(\frac{\pi}{4}\) of the difference between the top function and the bottom function: \(A = \int_0^{\frac{\pi}{4}} \left( \text{top function} - \text{bottom function} \right) \, dx\)
Write the integral explicitly using the functions identified in step 2. For example, if sin x is on top, then: \(A = \int_0^{\frac{\pi}{4}} (\sin x - x) \, dx\)
Evaluate the integral by finding the antiderivatives of sin x and x, then apply the Fundamental Theorem of Calculus by substituting the limits 0 and \(\frac{\pi}{4}\) to find the exact area.

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Key Concepts

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

Definite Integral for Area Calculation

The definite integral of a function over an interval gives the net area between the curve and the x-axis. To find the area between two curves, integrate the difference of the functions over the given interval. This method accounts for the region enclosed by the curves.
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Definition of the Definite Integral

Identifying the Upper and Lower Functions

When finding the area between two curves, it is essential to determine which function lies above the other within the interval. The area is found by integrating the upper function minus the lower function. This ensures the computed area is positive and accurate.
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Limits of Integration

The limits of integration define the interval over which the area is calculated. In this problem, the interval is from 0 to π/4. Correctly applying these bounds ensures the integral covers the exact region enclosed by the curves.
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Related Practice
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