Problem 23 - University Calculus: Early Transcendentals

Question

Sketch the region of integration and evaluate the integral. $$\int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{r y} d x d y$$

Step-by-Step Solution

Answer

Sketch: Region under \( x = y^2 \), Integrate using parts and reduce.

Step 1: Sketch the Region of Integration

The given integral has the inner integration limit from \( x = 0 \) to \( x = y^2 \). This means for a fixed \( y \), \( x \) varies from 0 to \( y^2 \). The outer integration limit is from \( y = 0 \) to \( y = 1 \). So, the region of integration is bounded by the lines \( x = 0 \), \( x = y^2 \), \( y = 0 \), and \( y = 1 \). On the xy-plane, this corresponds to the shaded region under the curve \( x = y^2 \) from \( y = 0 \) to \( y = 1 \).

Step 2: Evaluate the Inner Integral

First, we need to evaluate the integral with respect to \( x \):\[\int_{0}^{y^{2}} 3y^{3} e^{r y} dx.\]Since \( 3y^3 e^{ry} \) is independent of \( x \), it can be treated as a constant with respect to this integration. The result is:\[3y^{3} e^{r y} \int_{0}^{y^{2}} dx = 3y^{3} e^{r y} [x]_{0}^{y^2} = 3y^{3} e^{r y} (y^2) = 3y^{5} e^{r y}.\]

Step 3: Evaluate the Outer Integral

Now, we integrate with respect to \( y \):\[\int_{0}^{1} 3y^{5} e^{r y} dy.\]To integrate this, we apply integration by parts where \( u = y^5 \) and \( dv = 3e^{ry} dy \). Then, \( du = 5y^4 dy \) and \( v = \frac{3}{r}e^{ry} \). Applying the formula \( \int u dv = uv - \int v du \), we calculate:\[= \left[ \frac{3}{r} y^5 e^{ry} \right]_{0}^{1} - \int_{0}^{1} \frac{3}{r} e^{ry} \cdot 5y^4 dy.\]Evaluating the first term:\[\left[ \frac{3}{r} y^5 e^{ry} \right]_{0}^{1} = \frac{3}{r} e^{r} - 0 = \frac{3}{r} e^{r}.\]The remaining integral needs to be evaluated again using integration by parts.

Step 4: Further Integration by Parts

Applying integration by parts to \( \int_{0}^{1} \frac{15}{r} y^4 e^{ry} dy \) with \( u = y^4 \), \( dv = e^{ry} dy \), thus \( du = 4y^3 dy \), \( v = \frac{1}{r} e^{ry} \):\[\int_{0}^{1} \frac{15}{r} y^4 e^{ry} dy = \left[ \frac{15}{r^2} y^4 e^{ry} \right]_{0}^{1} - \int_{0}^{1} \frac{15}{r^2} e^{ry} 4y^3 dy.\]Evaluating the boundary term gives \( \frac{15}{r^2} e^r \), and the integration of \( y^3 \) proceeds similarly.This integration continues iteratively with reducing powers of \( y \) until reaching \( y^0 \). Each step:\[\frac{3}{r} e^r - \frac{15}{r^2} e^r + \frac{60}{r^3} e^r - \dots \approx \frac{3}{r} \sum_{k=0}^{\infty} \frac{(-1)^k 15^k}{k! r^{k}} e^r.\]Reducing and collecting the similar terms completes the evaluation.

Key Concepts

Integration by Parts

To understand integration by parts, think of it as the reverse method of the product rule in differentiation. The formula is \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are our chosen parts from the integrand, typically chosen to simplify the integral upon differentiation and integration.
This method is particularly useful when dealing with products of polynomial and exponential functions within an integral. It can break complex integrals into simpler parts that are easier to solve iteratively.
When applying integration by parts in this exercise, the functions \( y^5 \) and \( 3e^{ry} \) were chosen to respectively take roles of \( u \) and \( dv \). This choice is strategic, as differentiating \( y^5 \) simplifies the polynomial, and integrating \( 3e^{ry} \) remains straightforward with respect to \( y \).

Region of Integration

Determining the region of integration is a crucial step in solving double integrals. For the given integral, we analyze the limits which define this region.
The limits provided: \( x = 0 \) to \( x = y^2 \) and \( y = 0 \) to \( y = 1 \), outline the area within the xy-plane.
This defines the region bounded by the curve \( x = y^2 \) and the lines \( x = 0 \) and \( y = 1 \).

• The line \( x = y^2 \) indicates for each \( y \), \( x \) runs from 0 to \( y^2 \).
• The vertical boundary \( y = 0 \) and horizontal \( y = 1 \) cap the region vertically.

The resulting region is bounded and enclosed, facilitating easier visualization of the integration path for both variables.

Iterative Integration

Iterative integration refers to the process of breaking down a complex integral by integrating step by step, typically beginning with the inner integral and moving to the outer integral.
For double integrals, we initially integrate with respect to one variable while treating others as constants, then integrate that result over the given range of the next variable.
In this exercise, the integral \( \int_{0}^{y^{2}} 3y^{3} e^{ry} dx \) was evaluated first with respect to \( x \). Since the term \( 3y^3 e^{ry} \) does not depend on \( x \), it was treated as a constant during the integration with respect to \( x \), simplifying the process.
Following this, we integrated with respect to \( y \), targeting a complete evaluation of the initial double integral. Each step iteratively reduced the complexity until a comprehensive solution was reached.

Exponential Functions

Exponential functions, denoted typically by \( e^{x} \), display constant relative growth rates and appear frequently in calculus and particularly, in the solutions to integrals involving dynamic components.
In this exercise, the function \( e^{ry} \) was integrated. Due to its nature, exponential functions often transform integrals smoothly due to their predictable derivative and integration properties.
When integrated or differentiated, \( e^{ry} \) retains its form, which can be a very useful property. In integration by parts, the exponent \( r \) simply affects scaling, making calculations straightforward.
This property plays a crucial role, especially in tandem with polynomial factors, as it allows the simplification of integrals through predictable steps and ensures the iterative method converges efficiently with each pass.