Problem 36 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

AREA BETWEEN CURVES In Exercises 31 through 38 , sketch the indicated region \(R\) and find its area by integration. \(R\) is the region bounded by the curves \(y=\frac{8}{x}\) and \(y=\sqrt{x}\) and the line \(x=8 .\)

Step-by-Step Solution

Answer

The area is approximately 4.21 square units.

Step 1 - Identify the Intersection Points

Find the points where the curves intersect by setting \( \frac{8}{x} = \sqrt{x} \) and solving for x. Squaring both sides gives \[ \left(\frac{8}{x}\right)^2 = (\sqrt{x})^2 \] \ => \ \frac{64}{x^2} = x \ => \ 64 = x^3 \ => \ x = 4. So, the intersection points are at (4, 2).

Step 2 - Set Up the Integral

Integrate the difference between the two functions from the intersection point to the given limit. The integral for the area R is given by: \[ \int_{4}^{8} \left( \sqrt{x} - \frac{8}{x} \right) dx \]

Step 3 - Integrate \(\sqrt{x}\)

Find the antiderivative of \(\sqrt{x}\): \[ \int \sqrt{x} \,dx = \int x^{1/2} \,dx = \frac{2}{3} x^{3/2} \]

Step 4 - Integrate \(\frac{8}{x}\)

Find the antiderivative of \(\frac{8}{x}\): \[ \int \frac{8}{x} \,dx = 8 \int \frac{1}{x} \,dx = 8 \ln|x| \]

Step 5 - Evaluate the Integrals

Evaluate the integrals from the intersection point (x=4) to the line (x=8): \[ \left[ \frac{2}{3} x^{3/2} - 8 \ln|x| \right]_{4}^{8} \] \ => \ \[ \left( \frac{2}{3} (8)^{3/2} - 8 \ln|8| \right) - \left( \frac{2}{3} (4)^{3/2} - 8 \ln|4| \right) \]

Step 6 - Simplify the Expression

Calculate each term individually: \[ \frac{2}{3}(8)^{3/2} = \frac{2}{3} \times 22.6274 \ => \approx 15.085 \] \[ 8 \ln|8| = 8 \times 2.0794 \approx 16.6352 \] \[ \frac{2}{3}(4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \approx 5.333 \] \[ 8 \ln|4| = 8 \times 1.3863 \approx 11.0904 \]

Step 7 - Calculate the Final Area

Now, compute the difference: \[ (15.085 - 16.6352) - (5.333 - 11.0904) = -1.5502 + 5.7574 = 4.2072 \] The area of region R is approximately 4.2072 square units.

Key Concepts

Integration

Integration involves finding the area under a curve or between curves. To determine the area between two curves, we integrate the difference of their functions over a given interval. In this exercise, we are calculating the area bounded by the curves of the functions \( y = \frac{8}{x} \) and \( y = \sqrt{x} \) from the intersection point to a specified boundary, x = 8. We use the integral:

\[ \int_{4}^{8} \left( \sqrt{x} - \frac{8}{x} \right) dx \]

Here, we first need to find the intersection points by solving \( \frac{8}{x} = \sqrt{x} \). Integration helps combine the curves and handle complex area calculations by visualizing and summing up an infinite number of infinitely small areas cumulatively.

Antiderivatives

Finding the area involves calculating antiderivatives, or the reverse of differentiation. This process lets us determine the specific integral result.

For \( y = \sqrt{x} \), the antiderivative is found as follows: \ \[ \int \sqrt{x} \ dx = \int x^{1/2} dx = \frac{2}{3} x^{3/2} \ \]
Similarly, for \( y = \frac{8}{x} \): \ \[ \int \frac{8}{x} dx = 8 \int \frac{1}{x} dx = 8 \ln|x| \ \]
These antiderivatives are essential for calculating the definite integral and thus the area between curves. They provide a formula to evaluate the area precisely over the given bounds.

Bounded Region

The bounded region refers to the area enclosed between specific curves and boundaries. In our problem, it's the area between the curves \( y = \frac{8}{x} \) and \( y = \sqrt{x} \), from their intersection at x = 4 to x = 8.

We determine this by integrating the difference of the functions from the lower bound to the upper bound:
\[ \int_{4}^{8} \left( \sqrt{x} - \frac{8}{x} \right) dx \]

Here, the difference \( \sqrt{x} - \frac{8}{x} \) helps find the vertical 'slice' between the curves for any x within the bounds. Accumulating these slices gives the total area.

Intersection Points

Intersection points are where two curves meet, and they indicate the boundaries for integration. To find these points, set the given equations equal:

\(\frac{8}{x} = \sqrt{x} \).

Squaring both sides helps simplify and solve for x:
\ \[ \left(\frac{8}{x} \right)^{2} = (\sqrt{x})^{2} \ \ \rightarrow \ \ \frac{64}{x^{2}} = x \ \] \ \[ \32x = 32xx^{3} = 64 \rightarrow x = 4 \ \]
We then substitute x = 4 back into either equation to get the y-coordinate, resulting in the point (4, 2).

These coordinates are crucial for setting up the integration limits; they define where the curves intersect and thus the bounds within which we calculate the area.