Problem 14 - Calculus and its applications

Question

Evaluate. $$ \int_{1}^{3} \int_{0}^{x} 2 e^{x^{2}} d y d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( e^9 - e \).

Step 1: Identify the Order of Integration

The given integral is a double integral with the inner integral being \( \int_{0}^{x} 2e^{x^2} \, dy \) and the outer integral as \( \int_{1}^{3} \, dx \). Identify that first, we must integrate with respect to \( y \) from 0 to \( x \), and then integrate the result with respect to \( x \) from 1 to 3.

Step 2: Integrate with Respect to y

The inner integral has limits from 0 to \( x \). Treat \( x \) as a constant since we are integrating with respect to \( y \). The antiderivative of 2 with respect to \( y \) is \( 2y \). Evaluating from 0 to \( x \), we find: \[ \int_{0}^{x} 2e^{x^2} \, dy = 2ye^{x^2} \Big|_{0}^{x} = 2xe^{x^2} - 0 = 2xe^{x^2}. \]

Step 3: Integrate the Result with Respect to x

Now, we integrate \( 2xe^{x^2} \) with respect to \( x \) using the fundamental theorem of calculus. To do this, apply a substitution: let \( u = x^2 \), so \( du = 2x \, dx \). The limits for \( u \) change from \( 1^2 = 1 \) to \( 3^2 = 9 \). The integral becomes: \[ \int_{1}^{3} 2xe^{x^2} \, dx = \int_{1}^{9} e^u \, du. \] The integral of \( e^u \) with respect to \( u \) is \( e^u \), giving: \[ e^u \Big|_{1}^{9} = e^9 - e^1. \]

Step 4: Calculate the Final Result

Substitute the limits into the expression from the previous step to get the final result. Evaluating gives: \[ e^9 - e. \]

Step 5: Final Answer

The final evaluation of the given double integral is \( e^9 - e \). This is an exact expression for the result.

Key Concepts

Multiple Integration

Double integrals are a type of multiple integral where we extend the idea of integration to functions of two variables. This involves calculating the volume under a surface over a certain region. It's like stacking up infinitely many sheets of paper, all with a tiny thickness (differential), and summing their areas.

In the context of the given exercise, we have a double integral that is nested:

• The inner integral is with respect to one variable, here it's y.
• The outer integral is with respect to another variable, here it's x.

Understanding the order of integration is crucial, as it defines the path of integration over the region, often simplified by tackling one variable at a time. This 2layering2 makes it essential to properly understand how each step impacts the total result.

Change of Variables

When integrating complex functions, sometimes a clever substitution can simplify the process. This strategy is known as the change of variables, or substitution method.

In our given task, when integrating the function with respect to x, we noticed the expression of the form x raised to the power. Here, performing substitution by letting u equal x squared simplified our integral significantly:

• We set u = x^2, which made du = 2xdx.
• This substitution converts the original function into a more manageable integral in terms of u.

Changes in limits are also necessary when utilizing substitution. Initially, our x-limits were from 1 to 3. Upon changing variables to u, these limits appropriately updated from 1² to 3².

Using change of variables is a powerful technique that transforms otherwise complex integrals into simpler forms, making them easier to solve.

Definite Integrals

A definite integral represents the evaluation of an integral with particular bounds, providing an exact numerical value. It's different from an indefinite integral, which represents a family of functions and includes a constant of integration.

In our exercise, the final goal was to evaluate the definite integral of the function over specified limits. After simplifying through integration and substitution, we calculated the final values:

• The inner integral transitioned from variable y to obtaining a function in x.
• The subsequent integration with respect to x, after substitution, allowed evaluation between the new lower and upper limits for u.

For our problem, this definite integration led to calculating the result as \( e^9 - e \). This precise evaluation tells us the total "volume" under the surface, which is an exact numeric solution for the problem given the specified region.