Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The evaluated integral is \( \frac{8}{3} \).

Step 1: Understand the Structure of the Integral

This is an iterated integral, meaning we perform integration sequentially for different variables. Start by integrating with respect to the inner variable first, which in this case is y.

Step 2: Integrate with respect to y

The inner integral is \( \int_{-1}^{1} \left|y^3\right| \, dy \). Since \( \left|y^3\right| = y^3 \) when \( y \geq 0 \) and \( \left|y^3\right| = -y^3 \) when \( y < 0 \), split the integral as follows:\[ \int_{-1}^{0} (-y^3) \, dy + \int_{0}^{1} (y^3) \, dy \].Calculate each part separately.

Step 3: Evaluate the Integral from -1 to 0

Calculate \( \int_{-1}^{0} (-y^3) \, dy \):1. The antiderivative of \(-y^3\) is \(-\frac{y^4}{4}\).2. Evaluate it from -1 to 0: - \( \left.-\frac{y^4}{4}\right|_{-1}^{0} = -\frac{0^4}{4} + \frac{(-1)^4}{4} = \frac{1}{4}\).

Step 4: Evaluate the Integral from 0 to 1

Calculate \( \int_{0}^{1} y^3 \, dy \):1. The antiderivative of \(y^3\) is \(\frac{y^4}{4}\).2. Evaluate it from 0 to 1: - \( \left.\frac{y^4}{4}\right|_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}\).

Step 5: Sum the Results of Step 3 and Step 4

Add the results of the two integrals from Step 3 and Step 4:\[ \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]This is the result of the integration with respect to \( y \).

Step 6: Integrate with respect to x

Now integrate the resulting expression from Step 5 over \(-2\) to \(2\) for \(x^2\):\[ \int_{-2}^{2} x^2 \times \frac{1}{2} \, dx = \frac{1}{2} \int_{-2}^{2} x^2 \, dx \].

Step 7: Evaluate the Integral with respect to x

Calculate \( \int_{-2}^{2} x^2 \, dx \):1. The antiderivative of \(x^2\) is \(\frac{x^3}{3}\).2. Evaluate it from \(-2\) to \(2\): - \( \left.\frac{x^3}{3}\right|_{-2}^{2} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3} \).

Step 8: Multiply the Result

Multiply \( \frac{16}{3} \) by \( \frac{1}{2} \) from Step 6:\[ \frac{1}{2} \times \frac{16}{3} = \frac{16}{6} = \frac{8}{3} \].

Step 9: Conclusion

The evaluated value of the given iterated integral is \( \frac{8}{3} \).

Key Concepts

Integration by Parts

Integration by Parts is a powerful method used to evaluate integrals, especially when the integrals have products of functions. It is derived from the product rule for differentiation and can be expressed using the formula:\[ \int u \, dv = uv - \int v \, du \]Here, \(u\) and \(dv\) are chosen from the original integrand. Successfully applying Integration by Parts requires an understanding of when and how to choose these parts. The goal is to transform a difficult integral into simpler parts.
In the context of our problem, although Integration by Parts is not directly employed, one should understand it as it might be a needed technique in different settings where more complex integrations occur.

Antiderivative Calculation

Antiderivative Calculation is the process of finding a function whose derivative is the given function. It is also known as indefinite integration. For example, for the function \(x^2\), the antiderivative is \(\frac{x^3}{3}\). In the solution of the iterated integral, finding these antiderivatives plays a crucial role, especially when evaluating the definite integrals.

• To find the antiderivative of \(-y^3\), notice that it's \(-\frac{y^4}{4}\).
• Similarly, the antiderivative of \(y^3\) is \(\frac{y^4}{4}\).

When these antiderivatives are evaluated over definite limits, they help simplify and eventually solve the integral.

Absolute Value Function

The Absolute Value Function is a mathematical function that returns the non-negative value of a number or expression. It's especially important when dealing with functions like \(|y^3|\) because the behavior changes depending on whether \(y\) is positive or negative.

• For \(y \geq 0\), \(|y^3| = y^3\).
• For \(y < 0\), \(|y^3| = -y^3\).

In the given problem, the use of absolute value necessitates splitting the integral over \([-1, 1]\) into two parts, \([-1, 0]\) and \([0, 1]\), and evaluating them separately. This helps in accurately calculating the integral without errors due to sign changes.

Definite Integral Evaluation

Definite Integral Evaluation involves calculating the net area under the curve of a function between two specified bounds. This involves substituting the bounds into the antiderivatives and finding the difference.

• Once the antiderivative is known, the difference between its value at the upper and lower limits gives the value of the definite integral.
• For instance, in the example \(\int_{0}^{1} y^3 \, dy\), first find \(\frac{y^4}{4}\), and then evaluate it from 0 to 1 to get \(\frac{1}{4}\).

By understanding how to evaluate definite integrals, one can determine the total value effectively, accounting for all subtleties like changes in intervals or possible discontinuities.