Question

Evaluate the integral and subsequent iterated integral. (a) \(\int_{2}^{3}\left(6 x^{2}+4 x y-3 y^{2}\right) d y\) (b) \(\int_{-3}^{-2} \int_{2}^{5}\left(6 x^{2}+4 x y-3 y^{2}\right) d y d x\)

Short answer

The evaluated iterated integral is -6.

Step-by-step solution

Evaluate the Inner Integral

Calculate the inner integral for part (a): \[ \int_{2}^{3} (6x^2 + 4xy - 3y^2) \, dy. \]Integrate each term separately:1. \( \int 6x^2 \, dy = 6x^2y \)2. \( \int 4xy \, dy = 2xy^2 \)3. \( \int -3y^2 \, dy = -y^3 \)Using the limits 2 and 3, evaluate:\[ 6x^2(3) + 2x(3)^2 - (3)^3 - (6x^2(2) + 2x(2)^2 - (2)^3). \]

Simplify the Evaluation

Apply the limits to each term from Step 1:1. At \(y = 3\), the expression becomes: \[ 18x^2 + 18x - 27. \]2. At \(y = 2\), the expression becomes: \[ 12x^2 + 8x - 8. \] Subtract the results:\[ (18x^2 + 18x - 27) - (12x^2 + 8x - 8) = 6x^2 + 10x - 19. \]

Set Up the Outer Integral (b)

Now evaluate the outer integral for \(b\):\[ \int_{-3}^{-2} (6x^2 + 10x - 19) \, dx. \]This is the result from part (a) and becomes the integrand for \(x\).

Evaluate the Outer Integral

Integrate each term separately in the expression from Step 3:1. \( \int 6x^2 \, dx = 2x^3 \)2. \( \int 10x \, dx = 5x^2 \)3. \( \int -19 \, dx = -19x \)Apply the limits from -3 to -2:At \(x = -2\):\[ 2(-2)^3 + 5(-2)^2 - 19(-2). \]At \(x = -3\):\[ 2(-3)^3 + 5(-3)^2 - 19(-3). \]

Calculate Final Result

Evaluate the results from Step 4 and subtract:1. At \(x = -2\), the expression evaluates to: \[ -16 + 20 + 38 = 42. \]2. At \(x = -3\), the expression evaluates to: \[ -54 + 45 + 57 = 48. \] Subtracting gives:\[ 42 - 48 = -6. \]

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematical analysis. It concerns itself with the concept of integration. In simple terms, integration is the process of finding the integral of a function, which is essentially the reverse operation of differentiation. This is useful in situations where we want to determine the area under a curve, the accumulated change, or even the total output over time.
In integral calculus, integrals can either be indefinite or definite. An indefinite integral has no specific bounds and is represented with the integral symbol \(\int\), where we're looking for a function whose derivative is the given function. On the other hand, a definite integral has upper and lower bounds, represented as \(\int_{a}^{b}\), where we're finding the exact accumulation between two levels.
This branch of mathematics gives tools to solve real-world problems involving accumulation and area calculations, among others. In the given exercise, we solve both an iterated integral with definite bounds and a composite function that leads us to evaluate a double integral.

Double Integrals

Double integrals extend the concept of integration to functions of two variables, usually over a rectangular area in the xy-plane. When we compute a double integral, we're essentially finding the volume under a surface.
The given exercise involves a double integral, where we have to integrate the function \(6x^2 + 4xy - 3y^2\) first with respect to \(y\), and then with respect to \(x\). This is known as iterated integration.
Iterated integrals involve solving one integral inside the bounds of another. In simpler terms, you evaluate one integral first, treating the other variable as a constant, and then proceed to the next. In the provided example, we first integrate with respect to \(y\) from 2 to 3, resulting in a function of \(x\).
Once that is solved, the result becomes the new integrand for the next stage, where we integrate with respect to \(x\) from -3 to -2. Double integrals are key in calculating quantities such as mass, charge, and probability when these quantities are distributed over an area.

Partial Integration

Partial integration isn't to be confused with the integration by parts technique, although the terms might sound similar. In iterative or partial integration, we focus on integrating one part of the function at a time.
For iterated integrals, partial integration is the process of performing an integral on one variable while considering others as constants initially. For example, in the exercise, when we first integrated with respect to \(y\), each \(x\) term was treated as a constant.
This technique allows us to simplify complex integrals by breaking them down into manageable parts. Once the inner integration is complete, the outer integration can treat the resultant expression as a whole new function.
Partial integration in iterated integrals provides a systematic way to deal with multiple variables and complex regions. It is particularly valuable in fields like physics, engineering, and probability where multi-variable systems and relationships need to be analyzed.