Problem 14 - Calculus for Biology and Medicine

Question

All the integrals in problem are improper and converge. Explain in each case why the integral is improper, and evaluate each integral. $$ \int_{-2}^{0} \frac{d x}{(x+1)^{1 / 3}} $$

Step-by-Step Solution

Answer

The improper integral converges, and its value is 0.

Step 1: Identifying the Improperness

The given integral \( \int_{-2}^{0} \frac{d x}{(x+1)^{1 / 3}} \) is improper because the integrand becomes undefined at \( x = -1 \), which is within the limits of integration. The denominator \( (x+1)^{1/3} \) becomes zero when \( x+1 = 0 \), i.e., at \( x = -1 \). This causes a vertical asymptote in the function.

Step 2: Splitting the Integral

Since there is a singularity at \( x = -1 \), split the integral at this point: \[\int_{-2}^{0} \frac{d x}{(x+1)^{1 / 3}} = \int_{-2}^{-1} \frac{d x}{(x+1)^{1 / 3}} + \int_{-1}^{0} \frac{d x}{(x+1)^{1 / 3}}.\]Each component is still improper, but more manageable to handle individually with limits.

Step 3: Evaluating the First Integral

Evaluate \( \int_{-2}^{-1} \frac{d x}{(x+1)^{1 / 3}} \) by taking the limit as \( b \to -1^- \):\[\int_{-2}^{-1} \frac{d x}{(x+1)^{1 / 3}} = \lim_{b \to -1^-} \int_{-2}^{b} \frac{d x}{(x+1)^{1 / 3}}.\]Find the antiderivative: let \( u = (x+1)^{1/3} \), then \( du = \frac{1}{3}(x+1)^{-2/3} dx \). After solving, the antiderivative becomes \( \frac{3}{2}(x+1)^{2/3} \). Calculate the integral from -2 to \( b \).

Step 4: Calculating the First Integral's Limit

Evaluate the limit:\[\int_{-2}^{b} \frac{d x}{(x+1)^{1 / 3}} = \left[ \frac{3}{2} (x+1)^{2/3} \right]_{-2}^{b} = \frac{3}{2}(b+1)^{2/3} - \frac{3}{2}(1)^{2/3}\].As \( b \to -1^- \), \( (b+1) \to 0 \), therefore, the limit is \( \frac{3}{2}(0) - \frac{3}{2} = -\frac{3}{2} \).

Step 5: Evaluating the Second Integral

Evaluate \( \int_{-1}^{0} \frac{d x}{(x+1)^{1 / 3}} \) by taking the limit as \( a \to -1^+ \):\[\int_{-1}^{0} \frac{d x}{(x+1)^{1 / 3}} = \lim_{a \to -1^+} \int_{a}^{0} \frac{d x}{(x+1)^{1 / 3}}.\]Use the same antiderivative to solve and evaluate it from \( a \) to 0.

Step 6: Calculating the Second Integral's Limit

Evaluate the limit:\[\int_{a}^{0} \frac{d x}{(x+1)^{1 / 3}} = \left[ \frac{3}{2} (x+1)^{2/3} \right]_{a}^{0} = \frac{3}{2}(1)^{2/3} - \frac{3}{2}(a+1)^{2/3}\].As \( a \to -1^+ \), \( (a+1) \to 0 \), so the limit is \( \frac{3}{2} - 0 = \frac{3}{2} \).

Step 7: Combining the Results

Add the results from both integrals:\[-\frac{3}{2} + \frac{3}{2} = 0.\]Thus, the value of the improper integral is 0.

Key Concepts

Convergence of Integrals

When dealing with improper integrals, one of the core concerns is determining whether they converge or diverge. Convergence refers to the behavior of the integral as the interval of integration stretches towards or includes a point of singularity, like here with the integral \( \int_{-2}^{0} \frac{d x}{(x+1)^{1 / 3}} \). It has a singularity at \( x = -1 \) because the denominator becomes zero at this point.

To evaluate such integrals, the method is to break them into parts where the integration can be managed separately, each treated as simpler improper integrals. These smaller integrals are then evaluated using limits. For example, splitting the integral at \( x = -1 \) allows us to handle each piece effectively by employing limits towards the singular points:

• The first integral: \( \int_{-2}^{-1} \frac{d x}{(x+1)^{1 / 3}} \) evaluates as \( x \) approaches \( -1 \) from the left.
• The second integral: \( \int_{-1}^{0} \frac{d x}{(x+1)^{1 / 3}} \) is considered from the right side of \( -1 \).

If both resulting values are finite after calculating these limits, then the original integral converges. If any part leads to an infinite value, the integral diverges.
In this case, evaluating both parts results in finite values that cancel each other out, indicating convergence with a total value of zero.

Antiderivative

Finding the antiderivative is essential to solving integrals, including improper ones. In the context of \( \int_{-2}^{0} \frac{d x}{(x+1)^{1 / 3}} \), we first need to determine a usable antiderivative for \( \frac{1}{(x+1)^{1/3}} \). One effective approach is substitution, which simplifies the problem by transforming it into a more recognizable form.

Using this technique, let \( u = (x+1)^{1/3} \). Then, we differentiate to express \( dx \) in terms of \( du \), leading us to solve the integral in terms of \( u \) instead of \( x \). The resulting antiderivative is:

• \( \frac{3}{2}(x+1)^{2/3} \).

Once the antiderivative is known, it allows us to apply the limits of integration to evaluate the integral over a given interval, setting the scene for solving with limits to address improperness.
With proper calculation, this substitutive antiderivative returns a value for each sub-interval, which we then process through limit evaluations to determine convergence or divergence of each segment.

Limits of Integration

Limits of integration play a critical role in addressing improper integrals. Improper integrals have invalid points within the limits due to singularities or endpoints that extend to infinity. In this exercise, the invalid point occurs where the denominator of the integrand becomes zero: \( x = -1 \).

To handle this, the integral is split at the singularity, transforming one problematic integral into the sum of two smaller improper integrals:

• \( \int_{-2}^{-1} \) evaluated in terms of a limit as \( b \rightarrow -1^- \).
• \( \int_{-1}^{0} \) evaluated using a limit as \( a \rightarrow -1^+ \).

Each segment requires evaluation through these limits, ensuring that the approach towards the singularity is accounted for properly.
This process ensures that the behavior around such points is handled mathematically, leading to meaningful conclusions about the integral's convergence. Evaluating each part yields results which can be summed to provide the overall behavior of the original integral over its specified range, accounting for its improper elements successfully.