Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals in Problems 1-14. \(\int_{1}^{3} \int_{-y}^{2 y} x e^{y^{3}} d x d y\)

Step-by-Step Solution

Answer

\(\frac{1}{2} (e^{27} - e)\)

Step 1: Identify the Integrals

In this problem, we need to evaluate the given iterated integral: \[ \int_{1}^{3} \int_{-y}^{2 y} x e^{y^{3}} \; dx \; dy\] This is an iterated integral where we first integrate with respect to \(x\) and then with respect to \(y\).

Step 2: Inner Integral with Respect to x

The inner integral is: \[ \int_{-y}^{2y} x e^{y^{3}} \; dx \] Here, \(e^{y^{3}}\) is a constant with respect to \(x\), so we can factor it out of the integral: \[ e^{y^{3}} \int_{-y}^{2y} x \; dx \] Now solve \(\int_{-y}^{2y} x \; dx\). The antiderivative of \(x\) is \(\frac{x^2}{2}\).

Step 3: Evaluate the Inner Integral

Substitute the limits of integration for \(x\):\[ e^{y^{3}} \left[ \frac{(2y)^2}{2} - \frac{(-y)^2}{2} \right]\]This simplifies to:\[ e^{y^{3}} \left[ 2y^2 - \frac{y^2}{2} \right]\]Combine like terms:\[ e^{y^{3}} \cdot \frac{3y^2}{2} \] So the result of the inner integral is \( \frac{3y^2}{2} e^{y^3} \).

Step 4: Outer Integral with Respect to y

Now integrate the result from the inner integration with respect to \(y\):\[ \int_{1}^{3} \frac{3y^2}{2} e^{y^3} \; dy \]Use substitution: let \(u = y^3\), then \(du = 3y^2 \; dy\). This simplifies the integral to:\[ \int_{1^3}^{3^3} \frac{1}{2} e^u \; du \]This is a straightforward integral of the exponential function.

Step 5: Evaluate the Exponential Integral

Integrate:\[ \frac{1}{2} \int_{1^3}^{3^3} e^u \; du = \frac{1}{2} \left[ e^u \right]_{1}^{27} \]Calculate the indefinite integral:\[ = \frac{1}{2} (e^{27} - e)\] Thus, the final result for the iterated integral is \( \frac{1}{2} (e^{27} - e) \).

Key Concepts

Integral Calculus

Integral calculus is an essential part of calculus that focuses on the concept of integration. It is vital for solving problems involving areas under curves, calculating volumes, and even understanding physical properties like distance and velocity. In our exercise, we dealt with an iterated integral, which is an integral within an integral.
These types of integrals are used in multivariable calculus to find volumes or solve more complex problems involving functions of several variables.

• **Iterated Integrals:** These are evaluated one after the other, usually starting from the inside integral followed by the outer integral.
• **Applications:** Commonly used in physics and engineering to compute quantities that depend on multiple variables.

Integral calculus not only requires understanding of basic calculus principles but also the ability to evaluate these integrals using a sequence of techniques.

Integration Techniques

Integration techniques are various methods used to solve integrals. Each technique is suited for different types of integrals and functions. The ability to choose the right technique is crucial in integral calculus.
For the integral of our problem, we utilized basic integration rules for polynomials and exponential functions, along with substitution for the outer integral.

• **Basic Antiderivatives:** Knowing the antiderivatives, like \(\int x \, dx = \frac{x^2}{2}\), helps in solving polynomial integrals.
• **Factoring Constants:** In the given exercise, \(e^{y^3}\) was factored out of the inner integral because it was not dependent on \(x\).
• **Combining Terms:** After integrating, results like \(2y^2 - \frac{y^2}{2}\) were simplified to \(\frac{3y^2}{2}\).

These techniques are vital for dealing with different types of integrals that may appear in mathematical or applied contexts.

Substitution Method

The substitution method is a powerful technique used in integration to simplify complex integrals by transforming them into a more manageable form. This method is especially effective when dealing with composite functions or differential equations.
In our exercise, substitution made evaluating the outer integral feasible.

• **Choosing the Substitution:** Letting \(u = y^3\) in the outer integral turned the variable into a linear form that matched the exponential function.
• **Finding \(du\):** Calculating \(du = 3y^2 \, dy\) provided the differential that neatly replaced the original variables.
• **Integrating the New Variable:** The transformed integral became \(\frac{1}{2}\int e^{u} \, du\), a standard form for solving exponential integrals.

The substitution method not only simplifies integration but also expands the toolbox of techniques available for tackling integrals that initially seem intractable.