Problem 4 - Advanced Engineering Mathematics

Question

Evaluate the given partial integral. $$ \int_{\sqrt{y}}^{y^{3}}\left(8 x^{3} y-4 x y^{2}\right) d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( 2y^{13} - 2y^8 \).

Step 1: Understand the Integral

We are given a definite integral with respect to \( x \), where \( y \) is a parameter. The limits of the integral are dependent on \( y \): they range from \( \sqrt{y} \) to \( y^3 \). Our task is to evaluate \( \int_{\sqrt{y}}^{y^{3}} (8x^3y - 4xy^2) \, dx \).

Step 2: Break Down the Integrand

The integrand \( 8x^3y - 4xy^2 \) is a polynomial expression in terms of \( x \). We will integrate this expression term-by-term: \( 8x^3y \) and \( -4xy^2 \).

Step 3: Integrate Each Term

For the first term: \( 8x^3y \), the integral is \( \int 8x^3y \, dx = 8y \int x^3 \, dx = 8y \cdot \frac{x^4}{4} = 2yx^4 \).For the second term: \( -4xy^2 \), the integral is \( \int -4xy^2 \, dx = -4y^2 \int x \, dx = -4y^2 \cdot \frac{x^2}{2} = -2y^2x^2 \).

Step 4: Combine the Integrated Terms

The combined integrated expression is \( 2yx^4 - 2y^2x^2 \). Now, we need to evaluate this expression from the limits \( \sqrt{y} \) to \( y^3 \).

Step 5: Evaluate the Integrated Expression at the Upper Limit

Substitute \( x = y^3 \) into the integrated expression: \[ 2y(y^3)^4 - 2y^2(y^3)^2 = 2yy^{12} - 2y^2y^6 = 2y^{13} - 2y^8. \]

Step 6: Evaluate the Integrated Expression at the Lower Limit

Substitute \( x = \sqrt{y} \) into the integrated expression: \[ 2y(\sqrt{y})^4 - 2y^2(\sqrt{y})^2 = 2y \cdot y^2 - 2y^2 \cdot y = 2y^3 - 2y^3 = 0. \]

Step 7: Calculate the Definite Integral

Subtract the evaluated lower limit from the evaluated upper limit:\( (2y^{13} - 2y^8) - 0 = 2y^{13} - 2y^8. \)

Key Concepts

Definite Integral

A definite integral represents the area under a curve within a specific interval. Unlike indefinite integrals, which represent the antiderivative of a function, definite integrals calculate an actual numerical value. This value corresponds to the total accumulation of quantities under the curve, between two limits of integration.
In our exercise, the integral of the function \(8x^3y - 4xy^2\) with respect to \(x\), bounded by the limits \(\sqrt{y}\) and \(y^3\), is a definite integral. These limits of integration are called **variable limits**, typically used when another variable, like \(y\), influences the boundaries.

• Evaluating definite integrals involves two main steps: finding the antiderivative (the indefinite integral) and calculating the difference between the antiderivative evaluated at the upper and lower limits.
• The result provides a specific value based on the area and position defined by boundaries, assisting in problems dealing with real-world measurements.

Polynomial Integration

Polynomial integration involves finding the antiderivative of polynomial expressions. These expressions consist of sums of powers of a variable, typically making them straightforward to integrate term-by-term.
In our example, the integrand \(8x^3y - 4xy^2\) is polynomial with respect to \(x\). This means we can integrate each term separately:

• For \(8x^3y\), treat \(y\) as a constant, yielding the integral as \(2y x^4\), using the power rule for integration \(\int x^n \, dx = \frac{x^{n+1}}{n+1}\).
• Similarly, for \(-4xy^2\), treat \(y^2\) as constant, giving \(-2y^2x^2\) after integration.

After integrating each term, you combine the results to obtain the full integrated expression.

Variable Limits

When dealing with integrals like \(\int_{\sqrt{y}}^{y^{3}} (8x^3y - 4xy^2) \, dx\), we encounter **variable limits**. These occur when the limits of integration are not fixed numbers but expressions involving variables.
Variable limits are common in problems where the range of integration depends on another parameter in the problem (such as \(y\) in this exercise).

• They require substitution of the limit expression back into the integrated function to evaluate the definite integral at those points.
• This process can complicate the evaluation, as each limit needs careful substitution to faithfully reflect the original expression influenced by these variables.

Such variable limits necessitate a clear understanding of how they affect the evaluation and interpretation of the integral.

Parameter Integration

Parameter integration involves treating some variables as parameters during the integration process. In the context of the given problem, \(y\) functions as a parameter because it influences both the integrand and the limits of integration, although the integration is performed with respect to \(x\).
With parameter integration, it is important to:

• Recognize which variable is a constant with respect to the one being integrated (in this example, \(y\) remains constant when integrating with respect to \(x\)).
• Manage how the parameter affects not only the integrand but also the final evaluations across the bounds set by that parameter.

By dealing with the parameter correctly, you can derive integrals that reflect complex real-world dependencies and relationships embedded in mathematical descriptions.