Question

Evaluate the improper integrals. Each of these integrals has an infinite discontinuity either at an endpoint or at an interior point of the interval.\(\int_{-27}^{1} \frac{d x}{x^{2 / 3}}\)

Short answer

The value of the integral is 12.

Step-by-step solution

Identifying the Discontinuity

The function \( f(x) = x^{-2/3} \) becomes undefined at \(x = 0\) because division by zero is not possible. This signifies a point of discontinuity within the interval \([-27, 1]\). The integral is improper due to this discontinuity at \(x = 0\).

Splitting the Integral at the Discontinuity

We split the given integral into two separate integrals at the point of discontinuity \(x = 0\): \[\int_{-27}^{1} \frac{dx}{x^{2/3}} = \int_{-27}^{0} \frac{dx}{x^{2/3}} + \int_{0}^{1} \frac{dx}{x^{2/3}}\]

Evaluating Each Improper Integral

Both integrals are improper since they involve division by zero. We address this by using limits. The first integral is:\[\lim_{a\to 0^-} \int_{-27}^{a} x^{-2/3} \, dx\]and the second integral is:\[\lim_{b\to 0^+} \int_{b}^{1} x^{-2/3} \, dx\]

Integral Calculation for \(x^{-2/3}\)

For a power function integral \( \int x^{n} \, dx \), the antiderivative is \( \frac{x^{n+1}}{n+1} + C \) for \(n eq -1\). Thus, the antiderivative of \(x^{-2/3}\) is:\[\frac{x^{1/3}}{1/3} = 3x^{1/3}\]

Evaluating Limits for \([-27, 0)\)

Using the antiderivative, compute:\[\lim_{a\to 0^-} \left[ 3x^{1/3} \right]_{-27}^{a} \Rightarrow 3(a^{1/3}) - 3(-27)^{1/3}\]Since \(a\to 0^-\), the expression becomes \[0 - (-9)\], which is 9.

Evaluating Limits for \((0, 1]\)

Compute the integral from \(0+\) to 1:\[\lim_{b\to 0^+} \left[ 3x^{1/3} \right]_{b}^{1} \Rightarrow 3(1^{1/3}) - 3(b^{1/3})\]Since \(b\to 0^+\), the expression becomes \[3(1) - 0\], which is 3.

Summing Both Results

The value of the original improper integral is the sum of results from the two sides: \[9 + 3 = 12\]

Key concepts

Infinite Discontinuity

An infinite discontinuity occurs when a function grows without bound or becomes undefined within an interval. For the integral \( \int_{-27}^{1} \frac{dx}{x^{2/3}} \), the function \( f(x) = x^{-2/3} \) is undefined at \( x = 0 \), creating an infinite discontinuity. This is because dividing by zero is not allowed, as it results in a mathematical expression that lacks meaning. It is crucial to identify these discontinuities when dealing with integrals because they affect how we can evaluate the integral.

In practical terms, infinite discontinuities can often be addressed by using limits, allowing us to redefine the problem in a way that can be solved. By splitting the integral at the point of discontinuity, \( x = 0 \), we can handle each part separately, making the problem more manageable in our calculations.

Point of Discontinuity

A point of discontinuity is where a function is not defined or not continuous. In our integral, the point \( x = 0 \) disrupts the continuity of the function \( f(x) = x^{-2/3} \) as it goes to infinity at that location. Identifying these points is crucial because they mark where the conventional rules for integration don't apply.

In the integral \( \int_{-27}^{1} \frac{dx}{x^{2/3}} \), the split occurs at this point of discontinuity, dividing the integral into two smaller, more manageable parts: \( \int_{-27}^{0} \frac{dx}{x^{2/3}} \) and \( \int_{0}^{1} \frac{dx}{x^{2/3}} \). Each of these separate integrals can then be evaluated using different limits that approach the point of discontinuity from either the left or the right, ensuring we can still find the value of the whole integral.

Limits

Limits are a mathematical tool used to understand the behavior of a function as the input approaches a certain value. They are essential in dealing with improper integrals, especially when there's a point of infinite discontinuity.

For the function \( f(x) = x^{-2/3} \) at \( x = 0 \), we can't evaluate the integral \( \int \frac{dx}{x^{2/3}} \) directly due to the discontinuity. Instead, we use limits to redefine the problem:

• For \( \int_{-27}^{0} \), we calculate \( \lim_{a\to 0^-} \int_{-27}^{a} x^{-2/3} \, dx \) to approach the discontinuity from the left.
• For \( \int_{0}^{1} \), we use \( \lim_{b\to 0^+} \int_{b}^{1} x^{-2/3} \, dx \) to approach it from the right.

This approach focuses on the values of the function near the discontinuity, allowing us to evaluate otherwise unsolvable integrals.

Power Function Integral

Understanding power function integrals is crucial for solving problems involving polynomial expressions or functions of the form \( x^n \). The power function integral formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) (where \( n eq -1 \)) is fundamental in calculus.

In this scenario, we are dealing with \( x^{-2/3} \). Applying the power function rule, we calculate its antiderivative: \( \frac{x^{1/3}}{1/3} = 3x^{1/3} \). This antiderivative helps us evaluate the limits on either side of the discontinuity:

• On \([-27, 0)\), we use \[ 3(a^{1/3}) - 3(-27)^{1/3} \] as \( a \to 0^- \) and find it equals 9.
• On \((0, 1]\), it's \[ 3(1) - 3(b^{1/3}) \] as \( b \to 0^+ \), resulting in 3.

By summing these results, we derive the total value of the improper integral, which is 12. This process shows how powerful and systematic calculus techniques are in addressing otherwise complex integrals.