Problem 59 - Calculus Early Transcendentals

Question

Area of a region between curves Find the area of the region bounded by the curves \(y=\frac{x^{2}}{x^{3}-3 x}\) and \(y=\frac{1}{x^{3}-3 x}\) on the interval \([2,4] .\)

Step-by-Step Solution

Answer

Answer: The area is \(\frac{1}{3}\ln26\).

Step 1: Find points of intersection

To find the points of intersection, we set the two functions equal to each other and solve for x: \(\frac{x^{2}}{x^{3}-3 x} = \frac{1}{x^{3}-3 x}\) To solve this equation, we cross-multiply and get: \(x^{2}(x^{3}-3x) = 1(x^{3}-3x)\) Simplify by expanding the left side: \(x^5-3x^3 = x^3-3x\) Combining terms and setting everything to one side, we get: \(x^5 - 4x^3 + 3x = 0\) This equation factors as: \(x(x^2-1)(x^2-3) = 0\) From this, we find that the points of intersection are \(x=1, x=-1\) and \(x=\sqrt{3}, x=-\sqrt{3}\). However, those points are outside the interval of \([2, 4]\) so they won't affect our calculations.

Step 2: Subtract the functions

Now, we find the difference between the functions: \(d(y) = \frac{x^{2}}{x^{3}-3 x} - \frac{1}{x^{3}-3 x}\) In order to subtract these expressions, we need a common denominator, which in this case is \((x^{3}-3x)\). Therefore, the difference is: \(d(y) = \frac{x^{2}-1}{x^{3}-3 x}\)

Step 3: Integrate the difference

Now, integrate the difference on the interval \([2, 4]\): \(\int_{2}^{4} \frac{x^{2}-1}{x^{3}-3 x} dx\) Using u-substitution, let \(u = x^3 - 3x\), then \(du = (3x^2-3)dx\). We also need to change the limits of integration accordingly: u(2) = 8-6=2; u(4) = 64-12 = 52. So our new integral is: \(\int_{2}^{52} \frac{1}{3(u)} du\) Now, we can integrate this function: \(\frac{1}{3} \ln|u| \Big|_2^{52} = \frac{1}{3} (\ln|52|-\ln|2|)\) Now, we have found the area enclosed by the curves is: \(A = \frac{1}{3} (\ln|52|-\ln|2|) = \frac{1}{3}\ln\frac{52}{2} = \frac{1}{3}\ln26\) Thus, the area of the region bounded by the curves \(y=\frac{x^{2}}{x^{3}-3 x}\) and \(y=\frac{1}{x^{3}-3 x}\), on the interval \([2,4]\), is \(\frac{1}{3}\ln26\).

Key Concepts

Area between curves

When calculating the area between two curves, you need to find the space that is bounded by these graphs in a particular interval. This requires determining where the curves intersect, which acts as the boundaries of the area you are interested in. For our exercise, the curves given are

• \( y = \frac{x^2}{x^3 - 3x} \)
and
• \( y = \frac{1}{x^3 - 3x} \)

and they are evaluated on the interval

• \([2,4]\)

. To find the area, first determine where these functions overlap within the specified interval. This involves setting the functions equal to each other to find the x-values where they cross. These intersections may occur outside your interval, but they establish regions where one function is above the other. The function that is higher at any point in the interval sets the upper boundary, while the lower one sets the lower boundary for calculating the area.

Definite integral

The next step is to use calculus to calculate the exact area. The technique involves definite integration, which allows you to calculate the accumulation of quantities over a specified interval.

• A definite integral is represented as \( \int_a^b f(x) \, dx \) where \(a\) and \(b\) are the limits of integration that define the interval.
• In this exercise, the difference \( f(x) = \frac{x^2 - 1}{x^3 - 3x} \) represents the height at each point, with the area being the sum of these heights over the interval \([2, 4]\).

By integrating this function over the interval, we can find the total area between the two curves. This will provide us with a numeric value that represents the space between the curves in the given interval.

u-substitution

Often in calculus, the functions you need to integrate can be quite complex. U-substitution is a handy technique for simplifying integration. It involves transforming the function into a form that is easier to integrate. For the exercise in question:

• We let \( u = x^3 - 3x \), the denominator of our function, which simplifies the expression.
• This changes the differential \( dx \) to \( du = (3x^2 - 3)dx \), allowing us to reframe our integral in terms of \( u \).

Once substitution is made, we adjust the limits of integration to match the new variable \( u \). Solving the integration in terms of \( u \), then substituting back to the original variable, provides the solution to the area between the curves over the particular interval.

Integration techniques

Calculating areas between curves often requires a range of integration techniques beyond simple integration. In our exercise, besides U-substitution, we also leverage properties of logarithms and algebraic simplification.

• After finding an integral like \( \int \frac{1}{3u} \, du \), basic integrals are used to obtain solutions such as \( \frac{1}{3} \ln|u| \).
• Finally, evaluate the definite integral by substituting the limits into the log expression to find the difference at the bounds.
• Remember, integration is not only about doing mechanical steps but understanding the properties and transformations possible to arrive at a solution more efficiently.

These techniques help break down complex expressions into manageable parts so that calculating the area becomes straightforward.