An area function Consider the functions y = x²/a and y = √x/a, where a\u003e0. Find A(a), the area of the region between the curves.

Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 6 - Applications of Integration

문제 6.R.41

6장, 문제 6.R.41

An area function Consider the functions y = x²/a and y = √x/a, where a>0. Find A(a), the area of the region between the curves.

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Identify the two functions given: \( y = \frac{x^2}{a} \) and \( y = \frac{\sqrt{x}}{a} \), where \( a > 0 \).

Determine the points of intersection by setting the two functions equal: \( \frac{x^2}{a} = \frac{\sqrt{x}}{a} \). Simplify and solve for \( x \) to find the limits of integration.

Set up the integral for the area \( A(a) \) between the curves. Since \( y = \frac{\sqrt{x}}{a} \) is above \( y = \frac{x^2}{a} \) in the interval between the points of intersection, the area is given by the integral \( A(a) = \int_{x_1}^{x_2} \left( \frac{\sqrt{x}}{a} - \frac{x^2}{a} \right) \, dx \), where \( x_1 \) and \( x_2 \) are the intersection points.

Factor out \( \frac{1}{a} \) from the integral to simplify: \( A(a) = \frac{1}{a} \int_{x_1}^{x_2} \left( \sqrt{x} - x^2 \right) \, dx \).

Evaluate the integral by integrating each term separately: \( \int \sqrt{x} \, dx = \int x^{1/2} \, dx \) and \( \int x^2 \, dx \). Then substitute the limits \( x_1 \) and \( x_2 \) to express \( A(a) \) in terms of \( a \).

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Definite Integrals and Area Between Curves

The area between two curves over an interval is found by integrating the difference of the functions. Specifically, the integral of the upper function minus the lower function with respect to x gives the area between them. This requires identifying the correct limits of integration where the curves intersect.

Finding Points of Intersection

To determine the limits of integration, solve for x where the two functions are equal. These intersection points define the interval over which the area between the curves is calculated. Setting y = x²/a equal to y = √x/a and solving for x is essential.

Handling Parameters in Functions

The parameter 'a' affects the shape and position of the curves. When finding the area A(a), it is important to treat 'a' as a positive constant throughout the integration and algebraic manipulation. This ensures the final expression for area is correctly expressed in terms of 'a'.