Problem 66 - Calculus Single Variable

Question

In each of Exercises \(58-69\) use the Comparison Theorem to determine whether the given improper integral is convergent or divergent. In some cases, you may have to break up the integration before applying the Comparison Theorem. \(\int_{0}^{1} x^{-1 / 2}(1-x)^{-3 / 4} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{0}^{1} x^{-1/2} (1-x)^{-3/4} \, dx \) is convergent.

Step 1: Identify the Function for Comparison

The given integral is \( \int_{0}^{1} x^{-1 / 2} (1-x)^{-3 / 4} \, dx \). To apply the Comparison Theorem, let's first examine the behavior of the integrand near the limits of integration. As \( x \to 0^{+} \), the function behaves as \( x^{-1/2} \), and near \( x \to 1^{-} \), the function behaves as \((1 - x)^{-3/4}\). We will need separate comparisons near the endpoints.

Step 2: Compare Near x = 0

For \( x \to 0^{+} \), consider the integral \( \int_{0}^{1} x^{-1/2} \, dx \). This can be evaluated as \( \int_{0}^{1} x^{p} \, dx \) where \( p = -1/2 \). Since \( p > -1 \), the integral \( \int_{0}^{1} x^{-1/2} \, dx \) is convergent. Hence, near \( x = 0 \), the original integral is bounded by a convergent integral.

Step 3: Compare Near x = 1

For \( x \to 1^{-} \), consider the integral \( \int_{0}^{1} (1-x)^{-3/4} \, dx \). Here, this is equivalent to \( \int_{0}^{1} t^{-3/4} \, dt \) under the substitution \( t = 1-x \), which becomes \( \int_{0}^{1} t^{-3/4} \, dt \). Since \(-3/4 > -1\), this integral also converges.

Step 4: Conclusion Using the Comparison Theorem

Both comparison integrals are convergent, meaning that for \( x \to 0^{+} \) and \( x \to 1^{-} \), the integrand of the original integral is dominated by a function whose integral is known to converge on \([0, 1]\). By the Comparison Theorem, the given integral \( \int_{0}^{1} x^{-1/2} (1-x)^{-3/4} \, dx \) converges as well.

Key Concepts

Improper Integrals

Improper integrals arise when the interval of integration is infinite or when the integrand becomes infinite within the interval of integration. In simpler terms, improper integrals deal with those cases where standard definite integrals struggle due to infinite discontinuities or unbounded intervals. For instance, the integral \( \int_{a}^{b} f(x) \, dx \) becomes improper if:\

• The interval \( [a, b] \) includes points at which the integrand \( f(x) \) becomes undefined or approaches infinity.
• Either or both \( a \) and \( b \) are infinite, extending the interval beyond typical finite bounds.

In our exercise, the function \( x^{-1/2}(1-x)^{-3/4} \) creates potential problems at the endpoints \( x = 0 \) and \( x = 1 \) because both expressions approach infinity. Thus, even though the integral is over a finite interval \([0,1]\), the behavior of the function at \( x = 0 \) and \( x = 1 \) makes it improper.

Convergence and Divergence

When tackling improper integrals, we need to determine whether they converge or diverge. Convergence means the integral has a finite value, while divergence implies it's infinite or undefined. This distinction is crucial because it tells us if a function's integral provides a meaningful value or not. Let's see how it works:\

• Convergence: If the integral of a function \( f(x) \) from \( a \) to \( b \) results in a finite value, we say the integral converges. This suggests the area under the curve \( f(x) \) across the specified interval is finite.
• Divergence: An integral will diverge if the accumulated area turns out to be infinite or behaves wildly at any point in the interval, indicating the lack of a finite integral value.

In the example given, we determined convergence using the Comparison Theorem. We compared our target integral with two simpler integrals for \( x \to 0^{+} \) and \( x \to 1^{-} \). Both were convergent, thus proving our original integral's convergence by domination with functions of known behavior.

Calculus Techniques

To determine convergence or divergence, calculus offers us several powerful techniques. One such method is the Comparison Theorem, widely used for improper integrals. The basic idea is to compare the troublesome integral with a simpler, known integral that we already analyzed. Here's how you can apply it effectively:\

• Identify and separate the points of difficulty within the integral's interval. Examine the behavior of the integrand near these troublesome points.
• Choose comparison functions that resemble the original function's behavior at these points but are easier to integrate.
• Evaluate the comparison integrals. If these integrals are convergent, and the original function's behavior is less severe or about the same, you can conclude convergence by the Comparison Theorem.

In our problem, we looked at the behavior near \( x = 0 \) with \( \int_{0}^{1} x^{-1/2} \, dx \) and \( x = 1 \) with \( \int_{0}^{1} (1-x)^{-3/4} \, dx \), both of which are simpler forms already known to converge. Hence, we used these existing insights to make an informed judgment on the more complex integral.