Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each iterated integral. $$ \int_{-2}^{2} \int_{-1}^{1}\left|x^{2} y^{3}\right| d y d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{8}{3} \).

Step 1: Analyze the inner integral

Consider the inner integral with respect to \( y \), which is \( \int_{-1}^{1} |x^2 y^3| \, dy \). Since \( x^2 \) is always non-negative, we can simplify this to \( x^2 \int_{-1}^{1} |y^3| \, dy \). Because \( |y^3| = y^3 \) when \( y \geq 0 \) and \( |y^3| = -y^3 \) when \( y < 0 \), we must split the integral at \( y = 0 \):\[\int_{-1}^{1} |y^3| \, dy = \int_{-1}^{0} -y^3 \, dy + \int_{0}^{1} y^3 \, dy\]

Step 2: Evaluate the inner integrals for negative and positive y

First, evaluate the integral from \(-1\) to \(0\):\[\int_{-1}^{0} -y^3 \, dy = \int_{-1}^{0} (-1)y^3 \, dy = -\left[ \frac{y^4}{4} \right]_{-1}^{0} = -\left( 0 - \left( \frac{(-1)^4}{4} \right) \right) = -\left( 0 - \frac{1}{4} \right) = \frac{1}{4}\]Next, evaluate the integral from \(0\) to \(1\):\[\int_{0}^{1} y^3 \, dy = \left[ \frac{y^4}{4} \right]_{0}^{1} = \frac{1}{4} - 0 = \frac{1}{4}\]

Step 3: Combine results of the inner integrals

Now add the results of the two inner integrals:\[\int_{-1}^{1} |y^3| \, dy = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]

Step 4: Evaluate the outer integral

Substitute the result of the inner integral back into the outer integral and evaluate it with respect to \( x \):\[\int_{-2}^{2} x^2 \cdot \frac{1}{2} \, dx = \frac{1}{2} \int_{-2}^{2} x^2 \, dx\]Evaluate \( \int_{-2}^{2} x^2 \, dx \):\[\int_{-2}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) = \left( \frac{8}{3} + \frac{8}{3} \right) = \frac{16}{3}\]

Step 5: Calculate the final result

Using the result of the outer integral evaluation:\[\int_{-2}^{2} x^2 \cdot \frac{1}{2} \, dx = \frac{1}{2} \times \frac{16}{3} = \frac{16}{6} = \frac{8}{3}\]

Step 6: Conclusion: Final Result

The value of the iterated integral \( \int_{-2}^{2} \int_{-1}^{1} |x^2 y^3| \, dy \, dx \) is \( \frac{8}{3} \).

Key Concepts

Absolute Value Integration

Integrating functions that involve absolute values can seem tricky at first. For example, when dealing with the inner integral \( \int_{-1}^{1} |x^2 y^3| \, dy \), we're faced with the absolute value of \( y^3 \). The absolute value function affects how we treat positive and negative inputs.
For \( y^3 \), \( |y^3| = y^3 \) when \( y \geq 0 \), and \( |y^3| = -y^3 \) when \( y < 0 \). This means that the function changes its expression at \( y = 0 \). Splitting our integral at this point allows us to address the behaviors on both sides:

• Negative interval: \( \int_{-1}^{0} -y^3 \, dy \)
• Positive interval: \( \int_{0}^{1} y^3 \, dy \)

By integrating in these segments, we evaluate each scenario separately, allowing us to correctly manage the absolute value function.

Splitting Integrals at Discontinuities

To manage absolute values effectively in integration, we often split the integral at points where the function may switch direction, which is sometimes referred to as a discontinuity point in the piecewise definition of the function. In our exercise, this happens at \( y = 0 \).
Doing so allows us to correctly handle different parts of the function that may not be simple enough to incorporate within a single sweep of integration. This method is powerful because:

• It ensures accuracy by allowing proper handling of changing function properties.
• Makes it feasible to integrate functions that might otherwise seem inaccessible due to complexity from absolute values.

In our case, by splitting the integral \( \int_{-1}^{1} |y^3| \, dy \) at \( y = 0 \), we take into account the different impacts of \( y \) being positive or negative, ensuring the correct treatment of \( y^3 \) on either side.

Definite Integration

Definite integration is the process of calculating the area under a curve within given limits. In iterated integrals, such as \( \int_{-2}^{2} x^2 \cdot \frac{1}{2} \, dx \), it helps us to find the precise value of the area, accounting for both the limits and the function behavior.
When working with definite integrals:

• You'll have upper and lower limits that tell you the range over which to integrate.
• The fundamental theorem of calculus is used to evaluate these integrals, transforming the indefinite integral into a tangible numerical result.

For example, once we solve the function \( \int_{-2}^{2} x^2 \, dx \), evaluating from \( -2 \) to \( 2 \) involves finding an antiderivative and calculating the difference between its values at these bounds. This process gives us \( \frac{16}{3} \), which is a huge step in our overall solution.

Polynomial Functions in Integration

Polynomial functions, like \( x^2 \) in our exercise, are among the simplest to handle regarding integration due to their simple structure and predictable derivatives. They are often encountered in integration tasks:

• Integration of polynomial functions follows a direct pattern: increase the power by one, then divide by the new power.
• For example, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) (where \( C \) is the constant of integration) is a general formula for polynomial functions.

In the context of definite integrals, constant terms drop out after subtraction given their equal evaluation at the limits. This property makes evaluating iterated integrals involving polynomial terms straightforward. In our exercise, this pattern made the outer integration of the function \( \int_{-2}^{2} x^2 \, dx \) result in a smooth calculation of \( \frac{16}{3} \), contributing significantly to arriving at our final answer of \( \frac{8}{3} \).