Question

Evaluate the iterated integrals. $$ \int_{1}^{3} \int_{-y}^{2 y} x e^{y^{3}} d x d y $$

Short answer

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

Step-by-step solution

Understanding the Problem

We are given an iterated integral \( \int_{1}^{3} \int_{-y}^{2y} x e^{y^{3}} \, dx \, dy \). This means we need to first integrate with respect to \(x\) while treating \(y\) as a constant, and then integrate the result with respect to \(y\).

Integrate with Respect to x

Consider the inner integral \( \int_{-y}^{2y} x e^{y^3} \, dx \). Here, \( e^{y^3} \) is a constant with respect to \( x \), so the integration becomes \( e^{y^3} \int_{-y}^{2y} x \, dx \). Evaluate the integral of \( x \) over \( x \):\[\int x \, dx = \frac{x^2}{2} + C\]From \(-y\) to \(2y\), we find:\[e^{y^3} \left[ \frac{(2y)^2}{2} - \frac{(-y)^2}{2} \right] = e^{y^3} \left[ 2y^2 - \frac{y^2}{2} \right] \]

Simplify the Expression

Simplify the expression inside the brackets:\[e^{y^3} \left[ 2y^2 - \frac{y^2}{2} \right] = e^{y^3} \left[ \frac{4y^2}{2} - \frac{y^2}{2} \right] = e^{y^3} \cdot \frac{3y^2}{2} = \frac{3y^2}{2}e^{y^3}\]

Integrate with Respect to y

Now, evaluate the outer integral \( \int_{1}^{3} \frac{3y^2}{2} e^{y^3} \, dy \). Notice that the integrand is a result of a substitution for \( e^{u} \) where \( u = y^3 \). Compute the derivative \( du = 3y^2 \, dy \), which matches exactly with our integrand's \( 3y^2 \, dy \). The integral then simplifies to:\[\frac{1}{2} \int e^{u} \, du\]Evaluate this from \( y = 1 \) to \( y = 3 \) so we translate into the limits for \( u \), \( u = 1^3 = 1 \) and \( u = 3^3 = 27 \):\[\frac{1}{2}\left[ e^{27} - e^{1} \right] = \frac{1}{2} \left[ e^{27} - e \right]\]

Final Result

After evaluating all steps, the final result of the iterated integrals is:\[\frac{1}{2} \left[ e^{27} - e \right]\]

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus, involving finding a function given its derivative. It is often viewed through the lens of the area under a curve, but there's much more to it. Different integration techniques exist to deal with different types of functions and integrals. For this exercise, we use iterated integrals, a type of repeated integration that is crucial in multivariable calculus. This technique involves solving integrals step-by-step, focusing on one variable at a time when the function has more than two variables. By using an iterated integral, we slowly peel away each layer of the function, simplifying as we go.
Here, knowing which technique to apply can vastly simplify complex calculations. Other techniques often used include substitution (changing variables to simplify the integral) and integration by parts (useful for products of functions). Each has its context and utility, making the study of integration rich and varied."},{

Substitution Method

The substitution method is akin to changing the perspective to simplify an integral. We introduce a new variable in place of a function of the original variables that appears inside the integral. This reduces the complexity and makes it easier to evaluate the integral. For example, in this problem, we dealt with the function \(e^{y^3}\), which appears repeatedly. By substituting \(u = y^3\), we express the integral in terms of \(u\) and its differential \(du\), greatly simplifying the integration process.

• Choose a substitution that simplifies the integral.
• Find the differential \(du\) of the substitution.
• Replace the original variable and differential in the integral.

Using substitution often involves translating the limits of integration, as shown when the limits \(y = 1\) and \(y = 3\) were translated to \(u = 1\) to \(u = 27\). This technique becomes second nature with practice, helping solve more complicated integrals efficiently."},{

Definite Integrals

Definite integrals provide a way to find the accumulated value of a function over a specific interval, which could be the total area under the curve or might represent a total quantity like mass or charge, depending on the context. In this iterated integral scenario, the definite integral allows us to evaluate its exact value over a specified region.
To evaluate a definite integral:

• Take the indefinite integral (the antiderivative) first.
• Apply the limits of integration to find the actual accumulated value.
• The result is a definitive number representing the total value.

For our problem, the entire process results in an expression that, after evaluation of our final limits, determines exactly the number \( \frac{1}{2}(e^{27} - e) \). This definite result is especially powerful in applications requiring exact values, such as physics and engineering."},{

Integration by Parts

Integration by parts is another vital technique, useful when integrating a product of functions. Derived from the product rule for differentiation, it provides a way to "distribute" an integral across terms, often simplifying the computation.

• Apply wherever there's a product of two functions.
• One function is differentiated, the other is integrated.
• Rearrange the original integral into a new form.

Although not directly applied in this particular problem, understanding integration by parts enriches your toolbox for tackling integrals where one function is best served by being differentiated, while the other integrates smoothly. It's analogous to applying the distributive rule in algebra, opening doors to solving problems that at first seem unsolvable."}]} ered ]}]} ermini facil ut