Question

Find the area of the region between the curves \(y=(x-8)^{-2 / 3}\) and \(y=0\) for \(0 \leq x<8\)

Short answer

The area is 6 square units.

Step-by-step solution

Understand the Problem

We need to find the area between the curve \(y = (x-8)^{-2/3}\) and the x-axis (\(y = 0\)) from \(x = 0\) to \(x < 8\). Since there is an asymptote at \(x = 8\), we will calculate the improper integral from 0 to 8.

Set Up the Integral

The integral to find the area is given by \[A = \int_0^8 (x-8)^{-2/3} \, dx\]This integral calculates the area between the curve and the x-axis over the given interval. It is an improper integral due to the infinity at \(x=8\).

Perform the Substitution

Let \(u = x-8\). Then \(du = dx\) and when \(x = 0\), \(u = -8\), and when \(x\) approaches 8, \(u\) approaches 0. Rewrite the integral in terms of \(u\):\[A = \int_{-8}^{0} u^{-2/3} \, du\]

Solve the Improper Integral

We now solve the integral:\[\int u^{-2/3} \, du = \frac{u^{-2/3+1}}{-2/3+1} = \frac{u^{1/3}}{1/3} = 3u^{1/3}\]Evaluate this from \(-8\) to \(0\).

Evaluate the Integral at Bounds

Evaluate \[3u^{1/3} \big|_{-8}^{0} = 3(0^{1/3}) - 3((-8)^{1/3})\]Since \(0^{1/3} = 0\) and \((-8)^{1/3} = -2\) (since the cube root of -8 is -2), the expression simplifies to:\[0 - (-6) = 6\]

Conclude the Area Calculation

The area of the region described is finite even though the integral is improper. The area between the curve \(y = (x-8)^{-2/3}\) and \(y = 0\) from \(x = 0\) to \(x < 8\) is 6 square units.

Key concepts

Improper Integral

In calculus, an improper integral is used to find areas or other quantities, even when the function or the limits of integration are not bounded. In this exercise, we face a specific type of improper integral. The equation given is \(y = (x-8)^{-2/3}\), and as \(x\) approaches 8, the function becomes undefined because it leads towards infinity. Nevertheless, we can still compute the area under the curve to some extent, allowing us to deal with situations where certain parts of the integral might be infinite.

When handling improper integrals, we modify our approach slightly. Instead of evaluating directly to the point of infinity (or where a function is undefined), we take a limit. We solve the integral up to a certain point and observe the behavior of the function. The key is understanding that even if the curve reaches an infinite value, strategic mathematical approaches can "tame" the problem through limits and substitutions, facilitating finite solutions in many cases. This process turns complex integrals into manageable solutions, as seen in this problem where the improper integral yields a finite area.

Area Between Curves

Finding the area between curves is a common problem-solving scenario in calculus. Essentially, you calculate the space trapped between two curve lines across a specified interval. In this case, we're interested in the space between the curve defined by \(y = (x-8)^{-2/3}\) and the x-axis \(y = 0\).

To find this area:

• Set up the integral across the range of \(x\) values you are interested in.
• Consider any transformations or simplifications needed (such as substitution) to make solving the integral easier.
• Account for any improper behavior by adopting the necessary calculus techniques.

In the given problem, handling the improper integral yields the actual area between the curve and the x-axis from \(x = 0\) to approaching 8, which is found to be 6 square units, despite the curve tending towards infinity as \(x\) approaches 8.

Integration by Substitution

Integration by substitution is a powerful technique used in calculus to make complex integrals more manageable. It is similar to changing variables in algebraic equations. Here, you replace a complicated part of the integrand with a single variable, simplifying the integral so it becomes easier to solve.

In our exercise:

• We substitute \(u = x - 8\), effectively shifting the axis to simplify the boundaries.
• The differential \(du = dx\) aids in seamlessly transforming the variables, ensuring we have correctly adjusted the integral limits. Thus, when \(x = 0\), \(u = -8\), and as \(x\) approaches 8, \(u\) approaches 0.
• This substitution makes the integral \(\int (x-8)^{-2/3} \, dx\) transform into \(\int u^{-2/3} \, du\).

Once substitution is complete, integration becomes straightforward. After evaluating the integral and substituting back, we find the evaluated bounds leading to an easily computable numerical result. Integration by substitution is beneficial for turning complex, cumbersome integrals into simpler forms, saving both time and effort in calculations.